The solver stopped because it reached a limit on the number of iterations or function evaluations before it minimized the objective to the requested tolerance. To proceed, try one or more of the following.
|1. Enable Iterative Display|
|2. Relax Tolerances|
|3. Start the Solver From Different Points|
|4. Check Objective and Constraint Function Definitions|
|5. Center and Scale Your Problem|
|6. Provide Gradient or Jacobian|
|7. Provide Hessian|
Display option to
This setting shows the results of the solver iterations.
To enable iterative display:
Using the Optimization app, choose Level
of display to be
with detailed message.
At the MATLAB® command line, enter
options = optimoptions('solvername','Display','iter');
Call the solver using the
For an example of iterative display, see Interpreting the Result.
What to Look For in Iterative Display
See if the objective function (
decreases. Decrease indicates progress.
Examine constraint violation (
to ensure that it decreases towards
See if the first-order optimality decreases towards
Decrease indicates progress.
See if the
Trust-region radius decreases
to a small value. This decrease indicates that the objective might
not be smooth.
What to Do
If the solver seemed to progress:
values larger than the defaults. You can see the default values in
the Optimization app, or in the Options table in the solver's function
Start the solver from its last calculated point.
If the solver is not progressing, try the other listed suggestions.
for example, are too small, the solver might not recognize when it
has reached a minimum; it can make futile iterations indefinitely.
To change tolerances using the Optimization app, use the Stopping criteria list at the top of the Options pane.
FiniteDifferenceStepSize option (or
can affect a solver's progress. These options control the step size
in finite differencing for derivative estimation.
For example, check that your objective and nonlinear constraint functions return the correct values at some points. See Check your Objective and Constraint Functions. Check that an infeasible point does not cause an error in your functions; see Iterations Can Violate Constraints.
Solvers run more reliably when each coordinate has about the same effect on the objective and constraint functions. Multiply your coordinate directions with appropriate scalars to equalize the effect of each coordinate. Add appropriate values to certain coordinates to equalize their size.
Example: Centering and Scaling. Consider minimizing
1e6*x(1)^2 + 1e-6*x(2)^2:
f = @(x) 10^6*x(1)^2 + 10^-6*x(2)^2;
f using the
opts = optimoptions('fminunc','Display','none','Algorithm','quasi-newton'); x = fminunc(f,[0.5;0.5],opts) x = 0 0.5000
The result is incorrect; poor scaling interfered with obtaining a good solution.
Scale the problem. Set
D = diag([1e-3,1e3]); fr = @(y) f(D*y); y = fminunc(fr, [0.5;0.5], opts) y = 0 0 % the correct answer
Similarly, poor centering can interfere with a solution.
fc = @(z)fr([z(1)-1e6;z(2)+1e6]); % poor centering z = fminunc(fc,[.5 .5],opts) z = 1.0e+005 * 10.0000 -10.0000 % looks good, but... z - [1e6 -1e6] % checking how close z is to 1e6 ans = -0.0071 0.0078 % reveals a distance fcc = @(w)fc([w(1)+1e6;w(2)-1e6]); % centered w = fminunc(fcc,[.5 .5],opts) w = 0 0 % the correct answer
If you do not provide gradients or Jacobians, solvers estimate gradients and Jacobians by finite differences. Therefore, providing these derivatives can save computational time, and can lead to increased accuracy.
For constrained problems, providing a gradient has another advantage.
A solver can reach a point
x such that
feasible, but finite differences around
lead to an infeasible point. In this case, a solver can fail or halt
prematurely. Providing a gradient allows a solver to proceed.
Provide gradients or Jacobians in the files for your objective function and nonlinear constraint functions. For details of the syntax, see Writing Scalar Objective Functions, Writing Vector and Matrix Objective Functions, and Nonlinear Constraints.
To check that your gradient or Jacobian function is correct,
CheckGradients option, as described in Checking Validity of Gradients or Jacobians.
If you have a Symbolic Math Toolbox™ license, you can calculate gradients and Hessians programmatically. For an example, see Symbolic Math Toolbox Calculates Gradients and Hessians.
For examples using gradients and Jacobians, see Minimization with Gradient and Hessian, Nonlinear Constraints with Gradients, Symbolic Math Toolbox Calculates Gradients and Hessians, Nonlinear Equations with Analytic Jacobian, and Nonlinear Equations with Jacobian.
Solvers often run more reliably and with fewer iterations when you supply a Hessian.
The following solvers and algorithms accept Hessians:
Write the Hessian as a separate function. For an example, see fmincon Interior-Point Algorithm with Analytic Hessian.
Give the Hessian as the third output of the objective function. For
an example, see Minimization with Dense Structured Hessian, Linear Equalities.
Give the Hessian as the third output of the objective function. For
an example, see Minimization with Gradient and Hessian.
If you have a Symbolic Math Toolbox license, you can calculate gradients and Hessians programmatically. For an example, see Symbolic Math Toolbox Calculates Gradients and Hessians.
Usually, you get this result because the solver was unable to
find a point satisfying all constraints to within the
However, the solver might have located or started at a feasible point,
and converged to an infeasible point. If the solver lost feasibility,
see Solver Lost Feasibility. If
this result, see
quadprog Converges to an Infeasible Point
To proceed when the solver found no feasible point, try one or more of the following.
|1. Check Linear Constraints|
|2. Check Nonlinear Constraints|
Try finding a point that satisfies the bounds and linear constraints by solving a linear programming problem.
Define a linear programming problem with an objective function that is always zero:
f = zeros(size(x0)); % assumes x0 is the initial point
Solve the linear programming problem to see if there is a feasible point:
xnew = linprog(f,A,b,Aeq,beq,lb,ub);
If there is a feasible point
xnew as the initial point and rerun your original
If there is no feasible point, your problem is not well-formulated. Check the definitions of your bounds and linear constraints.
After ensuring that your bounds and linear constraints are feasible (contain a point satisfying all constraints), check your nonlinear constraints.
Set your objective function to zero:
Run your optimization with all constraints and with the zero
objective. If you find a feasible point
= xnew and rerun your original problem.
If you do not find a feasible point using a zero objective function, use the zero objective function with several initial points.
If you find a feasible point
x0 = xnew and rerun your original problem.
If you do not find a feasible point, try relaxing the constraints, discussed next.
Try relaxing your nonlinear inequality constraints, then tightening them.
Change the nonlinear constraint function
c-Δ, where Δ is a positive number.
This change makes your nonlinear constraints easier to satisfy.
Look for a feasible point for the new constraint function, using either your original objective function or the zero objective function.
If you find a feasible point,
Look for a feasible point for the new constraint function, starting at the previously found point.
If you do not find a feasible point, try increasing Δ and looking again.
If you find no feasible point, your problem might be truly infeasible, meaning that no solution exists. Check all your constraint definitions again.
If the solver started at a feasible point, but converged to an infeasible point, try the following techniques.
Try a different algorithm. The
are usually the most robust, so try one or both of them first.
Tighten the bounds. Give the highest
ub vectors that you can. This can help the
solver to maintain feasibility. The
obey bounds at every iteration, so tight bounds help throughout the
quadprogConverges to an Infeasible Point
Usually, you get this message because the linear constraints
are inconsistent, or are nearly singular. To check whether a feasible
point exists, create a linear programming problem with the same constraints
and with a zero objective function vector
f. Solve using the
options = optimoptions('linprog','Algorithm','dual-simplex'; x = linprog(f,A,b,Aeq,beq,lb,ub,options)
linprog finds no feasible point, then
your problem is truly infeasible.
linprog finds a feasible point, then
try a different
quadprog algorithm. Alternatively,
change some tolerances such as
solve the problem again.
The solver reached a point whose objective function was less than the objective limit tolerance.
Your problem might be truly unbounded. In other words, there is a sequence of points xi with
lim f(xi) = –∞.
and such that all the xi satisfy the problem constraints.
Check that your problem is formulated correctly. Solvers try to minimize objective functions; if you want a maximum, change your objective function to its negative. For an example, see Maximizing an Objective.
Try scaling or centering your problem. See Center and Scale Your Problem.
Relax the objective limit tolerance by using
reduce the value of the
fsolve can fail to solve an equation for
various reasons. Here are some suggestions for how to proceed:
Try Changing the Initial Point.
on an initial point. By giving it different initial points, you increase
the chances of success.
Check the definition of the equation to make sure
that it is smooth.
fsolve might fail to converge
for equations with discontinuous gradients, such as absolute value.
fail to converge for functions with discontinuities.
Check that the equation is "square," meaning equal dimensions for input and output (has the same number of unknowns as values of the equation).
Change tolerances, especially
If you attempt to get high accuracy by setting tolerances to very
fsolve can fail to converge. If
you set tolerances that are too high,
fail to solve an equation accurately.
Check the problem definition. Some problems have no
real solution, such as
x^2 + 1 = 0.