I am currently trying to solve a complementarity problem with a function that features a downward discontinuity, using the mcpsolve() function of the NLsolve package in Julia. The function is reproduced here for specific parameters, and the numbers below refer to the three panels of the figure.
Unfortunately, the algorithm does not always return the interior solution, even though it exists:
In (1), when starting at 0, the algorithm stays at 0, thinking that the boundary constraint binds,
In (2), when starting at 0, the algorithm stops right before the downward jump, even though the solution lies to the right of this point.
These problems occur regardless of the method used - trust region or Newton's method. Ideally, the algorithm would look for potential solutions in the entire set before stopping.
I was wondering if some of you had worked with similar functions, and had found a clever solution to bypass these issues. Note that
Starting to the right of the solution would not solve these problems, as they would also occur for other parametrization - see (3) this time,
It is not known a priori where in the parameter space the particular cases occur.
As an illustrative example, consider the following piece of code. Note that the function is smoother, and yet here as well the algorithm cannot find the solution.
function f!(x,fvec)
if x[1] <= 1.8
fvec[1] = 0.1 * (sin(3*x[1]) - 3)
else
fvec[1] = 0.1 * (x[1]^2 - 7)
end
end
NLsolve.mcpsolve(f!,[0.], [Inf], [0.], reformulation = :smooth, autodiff = true)
Once more, setting the initial value to something else than 0 would only postpone the problem. Also, as pointed out by Halirutan, fzero from the Roots package would probably work, but I'd rather use mcpsolve() as the problem is initially a complementarity problem.
Thank you very much in advance for your help.
Related
I am currently interested in billiards. However, I am interested in special billiards with a non-conventional reflection law and specific rules for the trajectories. I, therefore, need to calculate the trajectories using a differential equation solver. Finding one is not a problem at all. However, I still have trouble finding a suitable solution for the reflection. Previously I was working in Mathematica, whose Numerical ODE solver has a WhenEvent option:
Example:
NDSolve[{y''[t] == -9.81, y[0] == 5, y'[0] == 0, WhenEvent[y[t] == 0, y'[t] -> -0.95 y'[t]]}, y, {t, 0, 10}];
The solution that this line of code gives is a bouncing ball.
Basically after each integration step, it checks whether the condition is true and if so, it performs an action. (I suspect it checks if y has switched sign. If, for example, one puts in
WhenEvent[y²[t]==0],
the quantity does not switch sign and this method fails.)
Now, I would like to switch from Mathematica to something that is more openly available (C++ or python based.) but I could not find anything that can has this or a similar options. Does anyone perhaps have an Idea, what I could use instead? Basically I am looking for the option to check for a condition after each integration step and if the condition is met perform an action on the solution.
Does anyone have an idea what I could use?
Any help appreciated
I am trying to understand how to set the gap for Gurobi.Optimizer, because it solves too long when the default gap threshold is used (10e-4).
I couldn't find it anywhere (they might be referred to as attributes or parameters).
PS: Also, I am trying to find the right tutorial or instructions on how to use the solver in JuMP. I checked here https://juliahub.com/docs/Gurobi/do9v6/0.7.7, but they don't reveal the meanings of different attributes and inputs. Please, send me one in case somebody knows.
Best,
A.
You can set the MIP gap via the MIPGap parameter:
using JuMP, Gurobi
model = Model(Gurobi.Optimizer)
set_optimizer_attribute(model, "MIPGap", 0.1)
You can read more about JuMP here: https://github.com/jump-dev/Gurobi.jl
I'm new to SCIP, so I'm not sure if this is a bug or if I'm just doing something wrong.
I have a MIP instance that solves perfectly using SCIP, however when I try to solve a copy of the model SCIP says that it is infeasible. It seems to be more noticeable when presolve is turned off.
I'm using windows with the pre-built SCIP v3.2.0. The model only has binary and integer variables.
The following code outlines my attempt:
SCIP* _scip, subscip;
SCIPcreate(&_scip);
SCIPincludeDefaultPlugins(_scip);
SCIPcreateProbBasic(_scip, "interval_solver")); // create an empty problem
SCIPsetPresolving(_scip, SCIP_PARAMSETTING_OFF, true); //disable presolving
// build model (snipped)
SCIPsolve(_scip); // succeeds and gives feasible solution
SCIP_Bool valid = FALSE;
SCIPcreate(&subscip);
SCIPcopy(_scip, subscip, NULL, NULL, "1", TRUE, FALSE, TRUE, &valid);
SCIPsolve(subscip); // infeasible
Something that might be related (and seems weird to me) is that after solving the original problem (and getting a feasible solution), checking the solution reports an infeasible result. i.e.
SCIP_SOL* sol = SCIPgetBestSol(_scip);
SCIPcheckSol(_scip, sol, TRUE, TRUE, TRUE, TRUE, &valid);
gives:
solution value 1 violates bounds of <t_x71_(6,1275,6805)_(9,1275,6805)>[-0,0] by 1
Any ideas why this could be happening? Thanks!
Propagation in SCIP may take into account the best solution known so far and do reductions which are only valid for the problem of finding a solution better than this.
For example, if you have a minimization problem with n variables x_1,...,x_n with objective coefficients c_1,...,c_n >= 0 and already found a solution with x_1 = 1, x_2 = ... = x_n = 0, then propagation will globally fix x_1 to 0, because the objective of any solution with x_1 = 1 will be at least as large as the objective of the solution you already found.
This means that the solutions found so far may not be feasible anymore for the remaining problem (which looks for a strictly better solution).
In order to check the solution, you should check it in the original problem space, which you can do with SCIPcheckSolOrig().
Disabling presolving an propagation might help, but does not guarantee that the global presolved problem is not changed. Presolving in the LP solver should not be the problem, but might have changed the reported optimal LP solution (if there are multiple optima) and therefore caused a change in the solving process. This might have avoided your issue in this case, but probably by pure luck and the issue may still appear again on other instances. Moreover, the more features you disable, the more this will have a negative impact on your performance.
However, there is an easy solution to your problem: You can copy the original unchanged problem by using SCIPcopyOrig().
Some of the variable bounds were still being presolved. To fix the issue I needed to add:
SCIPsetBoolParam(_scip, "lp/presolving", FALSE);
This fixed most things, but the following also helped fix some 'check solution' issues:
SCIPsetIntParam(_scip, "propagating/maxrounds", 0);
SCIPsetIntParam(_scip, "propagating/maxroundsroot", 0);
I am trying to compute numerically the solutions for a system of many equations and variables (100+). I tried so far three things:
I now that the vector of p(i) (which contains most of the endogenous variables) is decreasing. Thus I gave simply some starting points, and then was increasing(decreasing) my guess when I saw that the specific p was too low(high). Of course this was always conditional on the other being fixed which is not the case. This should eventually work, but it is neither efficient, nor obvious that I reach a solution in finite time. It worked when reducing the system to 4-6 variables though.
I could create 100+ loops around each other and use bisection for each loop. This would eventually lead me to the solution, but take ages both to program (as I have no idea how to create n loops around each other without actually having to write the loops - which is also bad as I would like to increase/decrease the amount of variables easily) and to execute.
I was trying fminsearch, but as expected for that wast amount of variables - no way!
I would appreciate any ideas... Here is the code (this one the fminsearch I tried):
This is the run file:
clear all
clc
% parameter
z=1.2;
w=20;
lam=0.7;
tau=1;
N=1000;
t_min=1;
t_max=4;
M=6;
a_min=0.6;
a_max=0.8;
t=zeros(1,N);
alp=zeros(1,M);
p=zeros(1,M);
p_min=2;
p_max=1;
for i=1:N
t(i)= t_min + (i-1)*(t_max - t_min)/(N-1);
end
for i=1:M
alp(i)= a_min + (i-1)*(a_max - a_min)/(M-1);
p(i)= p_min + (i-1)*(p_max - p_min)/(M-1);
end
fun=#(p) david(p ,z,w,lam,tau,N,M,t,alp);
p0=p;
fminsearch(fun,p0)
And this is the program-file:
function crit=david(p, z,w,lam,tau,N,M,t,alp)
X = zeros(M,N);
pi = zeros(M,N);
C = zeros(1,N);
Xa=zeros(1,N);
Z=zeros(1,M);
rl=0.01;
rh=1.99;
EXD=140;
while (abs(EXD)>100)
r1=rl + 0.5*(rh-rl);
for i=1:M
for j=1:N
X(i,j)=min(w*(1+lam), (alp(i) * p(i) / r1)^(1/(1-alp(i))) * t(j)^((z-alp(i))/(1-alp(i))));
pi(i,j)=p(i) * t(j)^(z-alp(i)) * X(i,j)^(alp(i)) - r1*X(i,j);
end
end
[C,I] = max(pi);
Xa(1)=X(I(1),1);
for j=2:N
Xa(j)=X(I(j),j);
end
EXD=sum(Xa)- N*w;
if (abs(EXD)>100 && EXD>0)
rl=r1;
elseif (abs(EXD)>100 && EXD<0)
rh=r1;
end
end
Ya=zeros(M,N);
for j=1:N
Ya(I(j),j)=t(j)^(z-alp(I(j))) * X(I(j),j)^(alp(I(j)));
end
Yi=sum(Ya,2);
if (Yi(1)==0)
Z(1)=-50;
end
for j=2:M
if (Yi(j)==0)
Z(j)=-50;
else
Z(j)=(p(1)/p(j))^tau - Yi(j)/Yi(1);
end
end
zz=sum(abs(Z))
crit=(sum(abs(Z)));
First of all my recommendation: use your brain.
What do you know about the function, can you use a gradient approach, linearize the problem, or perhaps fix most of the variables? If not, think twice before you decide that you are really interested in all 100 variables and perhaps simplify the problem.
Now, if that is not possible read this:
If you found a way to quickly get a local optimum, you could simply wrap a loop around it to try different starting points and hope you will find a good optimum.
If you really need to make lots of loops (and a variable amount) I suppose it can be done with recursion, but it is not easily explained.
If you just quickly want to make a fixed number of loops inside each other this can easily be done in excel (hint: loop variables can be called t1,t2 ... )
If you really need to evaluate a function at a lot of points, probably creating all the points first using ndgrid and then evaluating them all at once is preferable. (Needless to say this will not be a nice solution for 100 nontrivial variables)
How'd I go about this one? I want to tween a value from one to another in x time. While also taking into account that it'd be nice to have an 'ease' at the start and end.
I know, I shouldn't ask really, but I've tried myself, and I'm stuck.
Please assume that to cause a delay, you need to call function wait(time).
One simple approach that might work for you is to interpolate along the unit circle:
To do this, you simply evaluate points along the circle, which ensures a fairly smooth movement, and ease-in as well as ease-out. You can control the speed of the interpolation by changing how quickly you alter the angle.
Assuming you're doing 1-dimensional interpolation (i.e. a simple scalar interpolation, like from 3.5 to 6.9 or whatever), it might be handy to use Y-values from -π/2 to π/2. These are given by the sine function, all you need to do is apply suitable scaling:
angle = -math.pi / 2
start = 3.5
end = 6.9
radius = (end - start) / 2
value = start + radius + radius * math.sin(angle)
I'm not 100% sure if this is legal Lua, didn't test it. If not, it's probably trivial to convert.
You may look at Tweener ActionScript library for inspiration.
For instance, you may borrow necessary equations from here.
If you need further help, please ask.