I am solving a standard LP problem:
min C'*x
S.t. A*x=b;x>=0;
After obtaining the solution through linprog, I want to obtain the optimal basis B corresponding to that solution. Simplex codes other than provided by MATLAB are very slow for the large-scale problems.
My problem has degeneracy.
The optimal basis of a non-degenerate LP is given by lambda = 0, where, lambda is the Lagrangian multipliers. Within MATLAB, lambda is available as the final output, i.e.
[x,fval,exitflag,output,lambda] = linprog(___)
So to find the basis, simply type k = find(lambda == 0).
However, the value of zero is problematic from a numerical perspective (almost nothing is ever completely 0 in floating-point arithmetic), so you might want to settle for something like k = find(lambda <= 1e-5). However, again, depending on the problem (and how well-behaved it is), this might not be correct either.
So, what can you do? There are basically two ways around this:
Use a commercial solver: commercial solvers tend to be much better in the accuracy of the Lagragian multipliers, especially for badly defined problems. Try Gurobi or CPLEX, and if you are a University student they are for free anyways. They have a similar way of getting lambda out, but are much more reliable in my experience.
Use the constraint values: you basically do k = find(x > 1e-5), and look what it gives you as results. This suffers from the same setbacks as using the Lagrangian, but it might help.
However, you then still need to deal with primal and dual degeneracy, if it occurs. Not to dive in too much, but basically you need to always check that you have exactly n active constraints (n being the number of optimization variables). If you have more or less than that, you have a problem, and you need to put an appropriate check into your code for that.
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I am trying to make a matlab program that optimize a flexure hinge designs. As i research matlab functions for multicriteria objectives i found multiple functions such as gamultiobj, fgoalattain and paretosearch however most of them outputed arrays of result instead of outputing 1 result. however i am looking for function that just output 1 single result for each variables. So i am trying to use fmincon function but since they only except single function to optimize. So i tried to look for ways to combine multiple objective criteria function. I have found a weighted sum method to combine it(for example f(x) = w1 * f1(x) + w2 * f2(x) ;) i have tried fmimax as well however it always weighted towards f1(first objective function in the function array) even though f2 can still be reduced. I am hoping to weight between those 2 objective funtion 50/50 compromise.
So basically i am just looking for functions or methods for nonlinear multicriteria objective with non linear constraint problem that when given functions that give single solution that each objective compromise so that none of the objectivebeing prioritize above others (aside from weighted sum method) ?
There are a few ways to approach this problem, but I suspect none will do exactly what you're looking for.
When you have more than one objective, and assuming the are competing, then there exists a trade-off. You need to decide the balance that you prefer between f1 and f2.
Multi-objective optimization, which you mention, provides you with a Pareto-optimal set of solutions. These solutions will be what is called non-dominated, no solution in the set will be better than another both in terms of f1 and f2. Each solution will represent a different trade-off between these values. It is up to you to look at the resulting set and decide which particular trade-off works best for your particular application.
As you've already found you can also do a weighted sum of objectives, converting this to a single-objective problem. However, this requires you do know your desired trade-off between the objectives and weight them accordingly. If you have some baseline solution you are trying to improve you might use that to normalize your function. For example, f = f1/f10 + f2/f20, would give an even balance to improving f1 and f2 relative to your initial design.
A third alternative is to convert one objective to a constraint. For example, if you will be happy with any solution that has f1 < c, for example, you can set this as a constraint and then use only f2 as an objective.
Each of these methods require that you determine the trade-off, or some satisfactory value, for f1 and f2. No optimization algorithms can come up with that balance for you.
I am studying Stochastic calculus, and occasionally we need to compute an integral (from -infinity to +infinity) for some complex distribution. In this case, it was
with the answer on the right. This is the code I put into Matlab (and I have the symbolic math toolbox), which Matlab simply cannot process:
>> syms x t
>> f = exp(1+2*x)*(1/((2*pi*t)^0.5))*exp(-(x^2)/(2*t))
>> int(f,-inf,inf)
ans =
-((2^(1/2)*pi^(1/2)*exp(2*t + 1)*limit(erf((2^(1/2)*((x*1i)/t - 2i))/(2*(-1/t)^(1/2))), x, -Inf)*1i)/(2*(-1/t)^(1/2)) - (2^(1/2)*pi^(1/2)*exp(2*t + 1)*limit(erf((2^(1/2)*((x*1i)/t - 2i))/(2*(-1/t)^(1/2))), x, Inf)*1i)/(2*(-1/t)^(1/2)))/(2*pi*t)^(1/2)
This answer at the end looks like nonsense, while Wolfram (via their free tool), gives me the answer that the picture above has. Am I missing something fundamental about doing such integrations in Matlab that the basic Mathworks pages don't cover? Where am I proceeding incorrectly?
In order to explain what is happening, we need some theory:
Symbolic systems such as Matlab or Mathematica calculate integrals symbolically by the Risch algorithm (yes, there is a method to mechanically calculate integrals, just like derivatives).
However, the Risch algorithms works differently than using derivation rules. Strictly spoken, it is not an algorithm but a semi-algorithm. This is, it is not deterministic one (as algorithms are).
This (semi) algorithm makes a series of transformations on the input expression (the one to be integrated), and in a specific point, it requires to ask if the transformed expression is equals to zero, because if it were zero, it cannot continue (the input is not integrable using a finite set of terms).
The problem (and the reason of the "semi-algoritmicity") is that, the (apparently simple) equation:
E = 0
Is indecidable (it is also called the constant problem). It means that there cannot exist a formal method to solve the constant problem, for any expression E. Of course, we know to solve the constant problem for specific forms of the expression E (i.e. polynomials), but it is impossible to solve the problem for the general case.
It also means that the Risch algorithm cannot be perfect (being able to solve any integral -integrable in finite terms-). In other words, the Risch algorithm will be as powerful as our ability to solve the constant problem for as many forms of the expression E as we can, but losing any hope of solving for the general case.
Different symbolic systems have similar, but different methods to try to solve equations (and therefore the constant problem), it explains why some of them can "solve" different sets of integrals than others.
And generalizing, because no symbolic system will never be able to solve the constant problem for the general case, it will neither be able to solve any integral (integrable in finite terms).
The second parameter of int() needs to be the variable you're integrating over (which looks like t in this case):
syms x t
f = exp(1+2*x)*(1/((2*pi*t)^0.5))*exp(-(x^2)/(2*t))
int(f,'t',-inf,inf) % <- integrate over t
I've got a problem with my equation that I try to solve numerically using both MATLAB and Symbolic Toolbox. I'm after several source pages of MATLAB help, picked up a few tricks and tried most of them, still without satisfying result.
My goal is to solve set of three non-polynomial equations with q1, q2 and q3 angles. Those variables represent joint angles in my industrial manipulator and what I'm trying to achieve is to solve inverse kinematics of this model. My set of equations looks like this: http://imgur.com/bU6XjNP
I'm solving it with
numeric::solve([z1,z2,z3], [q1=x1..x2,q2=x3..x4,q3=x5..x6], MultiSolutions)
Changing the xn constant according to my needs. Yet I still get some odd results, the q1 var is off by approximately 0.1 rad, q2 and q3 being off by ~0.01 rad. I don't have much experience with numeric solve, so I just need information, should it supposed to look like that?
And, if not, what valid option do you suggest I should take next? Maybe transforming this equation to polynomial, maybe using a different toolbox?
Or, if trying to do this in Matlab, how can you limit your solutions when using solve()? I'm thinking of an equivalent to Symbolic Toolbox's assume() and assumeAlso.
I would be grateful for your help.
The numerical solution of a system of nonlinear equations is generally taken as an iterative minimization process involving the minimization (i.e., finding the global minimum) of the norm of the difference of left and right hand sides of the equations. For example fsolve essentially uses Newton iterations. Those methods perform a "deterministic" optimization: they start from an initial guess and then move in the unknowns space essentially according to the opposite of the gradient until the solution is not found.
You then have two kinds of issues:
Local minima: the stopping rule of the iteration is related to the gradient of the functional. When the gradient becomes small, the iterations are stopped. But the gradient can become small in correspondence to local minima, besides the desired global one. When the initial guess is far from the actual solution, then you are stucked in a false solution.
Ill-conditioning: large variations of the unknowns can be reflected into large variations of the data. So, small numerical errors on data (for example, machine rounding) can lead to large variations of the unknowns.
Due to the above problems, the solution found by your numerical algorithm will be likely to differ (even relevantly) from the actual one.
I recommend that you make a consistency test by choosing a starting guess, for example when using fsolve, very close to the actual solution and verify that your final result is accurate. Then you will discover that, by making the initial guess more far away from the actual solution, your result will be likely to show some (even large) errors. Of course, the entity of the errors depend on the nature of the system of equations. In some lucky cases, those errors could keep also very small.
I am trying to solve a system of non linear equations using fsolve; lets say
F(x;lambda) = 0, where lambda is a vector of parameters, and x the vector I want to solve for.
I am using Matlab's fsolve.
I have 2 values of the parameter lambda, that I want to solve the system for. For the one value of lambda I get a solution, which seems alright.
For the other value of lambda I get a solution again (matlab exits with a flag of 1. However I know this is not an actual solution For example I know that some of the dimensions of x have to be equal to each other, and this is not the case in the solution I get from fsolve.
I have tried both trust-region and the levenberg-marquardt algorithm, and I am not getting any better results. (explicitly enforcing those x's to be the same, still seems to give solutions that are not consistent with what I would be expecting from the properties of the system)
My question is: do the algorithms used by fsolve depend on any kind of stability of the system? Could it be that changing the parameter lambda in the second case I mention above, I make the system unstable, and could that make fsolve having a hard time to solve it correctly?
Thank you, George
fsolve isn't "failing" - as commented by jucestain, it's giving you a local minimum, which is not necessarily a global minimum. This is what it's designed to do.
To improve your chances of obtaining a global minimum you need to either:
Know that your initial guess is good
Run the optimisation several times with a grid of initial guesses, and pick the best result
Add constraints to prevent the solver straying into areas you know to have local minima
Modify your cost function to remove local minima
If you ever come across a non-linear solver that can guarantee a global minimum, do let us know!
i want to solve this problem:
alt text http://img265.imageshack.us/img265/6598/greenshot20100727091025.png
i don't want to use "int", i want to use "quad" family (quad,dblquad,triplequad)
but i can't.
can you help me?
I assume that your real problem is more complex than this trivial one. The best solution is just to use a symbolic integral. Why is numerical integration difficult?
Numerical integration in ONE dimension typically requires on the order of say 100 function evaluations. (The exact number will be very dependent on the accuracy required, the limits, etc.) This makes a 2-d integral typically require on the order of 100^2 = 10000 function evals. So an adaptive, 5-d integral will require on the order of 100^5 = 1e10 function evaluations. (This is only a very rough order of magnitude estimate here.) My point is, you simply don't want to do that!
Better is to reduce the problem in complexity. If your integral is separable (as is this one) then do so! Reduce a 5-d problem into multiple 1-d problems.
Also, in many cases I see people wanting to do a numerical integration of a Gaussian PDF. See that this is easily solved using a call to erf or erfc, coupled with a transformation. The point is that in many cases special functions are defined to greatly reduce the complexity of a problem.
I should add that in many cases, the key to solving a difficult problem in mathematics is to use mathematics to reduce the problem to something simpler. If you can find a way to reduce the dimensionality of your problem just a bit, it will become much more tractable.
The integral you show is
Analytically solvable: always do analytically what you can
?equal to a number: constant expressions should be eliminated from numerical calculations
not easy to get calculated in MATLAB (or very correct).
You can use cumtrapz to integrate over each variable alone, and call trapz the final integration. Remember that this will blow up the error on any problem that is more complicated than the simple sum of linear functions.
Mathematica is more suited to nD integrations, if you have access to that.
matlab can do symbolic integration
>> x = sym('x'); y = sym('y'); z = sym('z'); u = sym('u'); v = sym('v');
>> int(int(int(int(int(x+y+z+u+v,1,5),-2,3),0,1),-1,1),0,1)
ans =
180
Just noticed you want to do numeric, not symbolic integration
If you look at the source of dblquad and triplequad
>> edit dblquad
you see that they just call the lower versions.
it should be possible for you to add a quadquad and a quintquad (or recursively an n-quad)