Optimization Toolbox 

This example shows how to use two nonlinear optimization solvers and how to set options. The nonlinear solvers that we use in this example are fminunc and fmincon.
All the principles outlined in this example apply to the other nonlinear solvers, such as fgoalattain, fminimax, lsqnonlin, lsqcurvefit, and fsolve.
The example starts with minimizing an objective function, then proceeds to minimize the same function with additional parameters. After that, the example shows how to minimize the objective function when there is a constraint, and finally shows how to get a more efficient and/or accurate solution by providing gradients or Hessian, or by changing some options.
Unconstrained Optimization Example
Consider the problem of finding a minimum of the function:
Plot the function to get an idea of where it is minimized
f = @(x,y) x.*exp(x.^2y.^2)+(x.^2+y.^2)/20; ezsurfc(f,[2,2])
The plot shows that the minimum is near the point (1/2,0).
Usually you define the objective function as a MATLAB file. For now, this function is simple enough to define as an anonymous function:
fun = @(x) f(x(1),x(2));
Take a guess at the solution:
x0 = [.5; 0];
Set optimization options to not use fminunc's default largescale algorithm, since that algorithm requires the objective function gradient to be provided:
options = optimoptions('fminunc','Algorithm','quasinewton');
View the iterations as the solver calculates:
options.Display = 'iter';
Call fminunc, an unconstrained nonlinear minimizer:
[x, fval, exitflag, output] = fminunc(fun,x0,options);
Firstorder Iteration Funccount f(x) Stepsize optimality 0 3 0.3769 0.339 1 6 0.379694 1 0.286 2 9 0.405023 1 0.0284 3 12 0.405233 1 0.00386 4 15 0.405237 1 3.17e05 5 18 0.405237 1 3.35e08 Local minimum found. Optimization completed because the size of the gradient is less than the default value of the function tolerance.
The solver found a solution at:
uncx = x
uncx = 0.6691 0.0000
The function value at the solution is:
uncf = fval
uncf = 0.4052
We will use the number of function evaluations as a measure of efficiency in this example. The total number of function evaluations is:
output.funcCount
ans = 18
Unconstrained Optimization with Additional Parameters
We will now pass extra parameters as additional arguments to the objective function. We show two different ways of doing this  using a MATLAB file, or using a nested function.
Consider the objective function from the previous section:
We parameterize the function with (a,b,c) in the following way:
This function is a shifted and scaled version of the original objective function.
Method 1: MATLAB file Function
Suppose we have a MATLAB file objective function called bowlpeakfun defined as:
type bowlpeakfun
function y = bowlpeakfun(x, a, b, c) %BOWLPEAKFUN Objective function for parameter passing in TUTDEMO. % Copyright 2008 The MathWorks, Inc. y = (x(1)a).*exp(((x(1)a).^2+(x(2)b).^2))+((x(1)a).^2+(x(2)b).^2)/c;
Define the parameters:
a = 2; b = 3; c = 10;
Create an anonymous function handle to the MATLAB file:
f = @(x)bowlpeakfun(x,a,b,c)
f = @(x)bowlpeakfun(x,a,b,c)
Call fminunc to find the minimum:
x0 = [.5; 0]; options = optimoptions('fminunc','Algorithm','quasinewton'); [x, fval] = fminunc(f,x0,options)
Local minimum found. Optimization completed because the size of the gradient is less than the default value of the function tolerance. x = 1.3639 3.0000 fval = 0.3840
Method 2: Nested Function
Consider the following function that implements the objective as a nested function
type nestedbowlpeak
function [x,fval] = nestedbowlpeak(a,b,c,x0,options) %NESTEDBOWLPEAK Nested function for parameter passing in TUTDEMO. % Copyright 2008 The MathWorks, Inc. [x,fval] = fminunc(@nestedfun,x0,options); function y = nestedfun(x) y = (x(1)a).*exp(((x(1)a).^2+(x(2)b).^2))+((x(1)a).^2+(x(2)b).^2)/c; end end
In this method, the parameters (a,b,c) are visible to the nested objective function called nestedfun. The outer function, nestedbowlpeak, calls fminunc and passes the objective function, nestedfun.
Define the parameters, initial guess, and options:
a = 2; b = 3; c = 10; x0 = [.5; 0]; options = optimoptions('fminunc','Algorithm','quasinewton');
Run the optimization:
[x,fval] = nestedbowlpeak(a,b,c,x0,options)
Local minimum found. Optimization completed because the size of the gradient is less than the default value of the function tolerance. x = 1.3639 3.0000 fval = 0.3840
You can see both methods produced identical answers, so use whichever one you find most convenient.
Constrained Optimization Example: Inequalities
Consider the above problem with a constraint:
The constraint set is the interior of a tilted ellipse. Look at the contours of the objective function plotted together with the tilted ellipse
f = @(x,y) x.*exp(x.^2y.^2)+(x.^2+y.^2)/20; g = @(x,y) x.*y/2+(x+2).^2+(y2).^2/22; ezplot(g,[6,0,1,7]) hold on ezcontour(f,[6,0,1,7]) plot(.9727,.4685,'ro'); legend('constraint','f contours','minimum'); hold off
The plot shows that the lowest value of the objective function within the ellipse occurs near the lower right part of the ellipse. We are about to calculate the minimum that was just plotted. Take a guess at the solution:
x0 = [2 1];
Set optimization options: use the interiorpoint algorithm, and turn on the display of results at each iteration:
options = optimoptions('fmincon','Algorithm','interiorpoint','Display','iter');
Solvers require that nonlinear constraint functions give two outputs: one for nonlinear inequalities, the second for nonlinear equalities. So we write the constraint using the deal function to give both outputs:
gfun = @(x) deal(g(x(1),x(2)),[]);
Call the nonlinear constrained solver. There are no linear equalities or inequalities or bounds, so pass [ ] for those arguments:
[x,fval,exitflag,output] = fmincon(fun,x0,[],[],[],[],[],[],gfun,options);
Firstorder Norm of Iter Fcount f(x) Feasibility optimality step 0 3 2.365241e01 0.000e+00 1.972e01 1 6 1.748504e01 0.000e+00 1.734e01 2.260e01 2 10 1.570560e01 0.000e+00 2.608e01 9.347e01 3 14 6.629160e02 0.000e+00 1.241e01 3.103e01 4 17 1.584082e01 0.000e+00 7.934e02 1.826e01 5 20 2.349124e01 0.000e+00 1.912e02 1.571e01 6 23 2.255299e01 0.000e+00 1.955e02 1.993e02 7 26 2.444225e01 0.000e+00 4.293e03 3.821e02 8 29 2.446931e01 0.000e+00 8.100e04 4.035e03 9 32 2.446933e01 0.000e+00 1.999e04 8.126e04 10 35 2.448531e01 0.000e+00 4.004e05 3.289e04 11 38 2.448927e01 0.000e+00 4.036e07 8.156e05 Local minimum found that satisfies the constraints. Optimization completed because the objective function is nondecreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
A solution to this problem has been found at:
x
x = 0.9727 0.4686
The function value at the solution is:
fval
fval = 0.2449
The total number of function evaluations was:
Fevals = output.funcCount
Fevals = 38
The inequality constraint is satisfied at the solution.
[c, ceq] = gfun(x)
c = 2.4608e06 ceq = []
Since c(x) is close to 0, the constraint is "active," meaning the constraint affects the solution. Recall the unconstrained solution was found at
uncx
uncx = 0.6691 0.0000
and the unconstrained objective function was found to be
uncf
uncf = 0.4052
The constraint moved the solution, and increased the objective by
fvaluncf
ans = 0.1603
Constrained Optimization Example: UserSupplied Gradients
Optimization problems can be solved more efficiently and accurately if gradients are supplied by the user. This example shows how this may be performed. We again solve the inequalityconstrained problem
To provide the gradient of f(x) to fmincon, we write the objective function in the form of a MATLAB file:
type onehump
function [f,gf] = onehump(x) % ONEHUMP Helper function for Tutorial for the Optimization Toolbox demo % Copyright 20082009 The MathWorks, Inc. r = x(1)^2 + x(2)^2; s = exp(r); f = x(1)*s+r/20; if nargout > 1 gf = [(12*x(1)^2)*s+x(1)/10; 2*x(1)*x(2)*s+x(2)/10]; end
The constraint and its gradient are contained in the MATLAB file tiltellipse:
type tiltellipse
function [c,ceq,gc,gceq] = tiltellipse(x) % TILTELLIPSE Helper function for Tutorial for the Optimization Toolbox demo % Copyright 20082009 The MathWorks, Inc. c = x(1)*x(2)/2 + (x(1)+2)^2 + (x(2)2)^2/2  2; ceq = []; if nargout > 2 gc = [x(2)/2+2*(x(1)+2); x(1)/2+x(2)2]; gceq = []; end
Make a guess at the solution:
x0 = [2; 1];
Set optimization options: we continue to use the same algorithm for comparison purposes.
options = optimoptions('fmincon','Algorithm','interiorpoint');
We also set options to use the gradient information in the objective and constraint functions. Note: these options MUST be turned on or the gradient information will be ignored.
options = optimoptions(options,'GradObj','on','GradConstr','on');
There should be fewer function counts this time, since fmincon does not need to estimate gradients using finite differences.
options.Display = 'iter';
Call the solver:
[x,fval,exitflag,output] = fmincon(@onehump,x0,[],[],[],[],[],[], ...
@tiltellipse,options);
Firstorder Norm of Iter Fcount f(x) Feasibility optimality step 0 1 2.365241e01 0.000e+00 1.972e01 1 2 1.748504e01 0.000e+00 1.734e01 2.260e01 2 4 1.570560e01 0.000e+00 2.608e01 9.347e01 3 6 6.629161e02 0.000e+00 1.241e01 3.103e01 4 7 1.584082e01 0.000e+00 7.934e02 1.826e01 5 8 2.349124e01 0.000e+00 1.912e02 1.571e01 6 9 2.255299e01 0.000e+00 1.955e02 1.993e02 7 10 2.444225e01 0.000e+00 4.293e03 3.821e02 8 11 2.446931e01 0.000e+00 8.100e04 4.035e03 9 12 2.446933e01 0.000e+00 1.999e04 8.126e04 10 13 2.448531e01 0.000e+00 4.004e05 3.289e04 11 14 2.448927e01 0.000e+00 4.036e07 8.156e05 Local minimum found that satisfies the constraints. Optimization completed because the objective function is nondecreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
fmincon estimated gradients well in the previous example, so the iterations in the current example are similar.
The solution to this problem has been found at:
xold = x
xold = 0.9727 0.4686
The function value at the solution is:
minfval = fval
minfval = 0.2449
The total number of function evaluations was:
Fgradevals = output.funcCount
Fgradevals = 14
Compare this to the number of function evaluations without gradients:
Fevals
Fevals = 38
Changing the Default Termination Tolerances
This time we solve the same constrained problem
more accurately by overriding the default termination criteria (options.TolX and options.TolFun). We continue to use gradients. The default values for fmincon's interiorpoint algorithm are options.TolX = 1e10, options.TolFun = 1e6.
Override two default termination criteria: termination tolerances on X and fval.
options = optimoptions(options,'TolX',1e15,'TolFun',1e8);
Call the solver:
[x,fval,exitflag,output] = fmincon(@onehump,x0,[],[],[],[],[],[], ...
@tiltellipse,options);
Firstorder Norm of Iter Fcount f(x) Feasibility optimality step 0 1 2.365241e01 0.000e+00 1.972e01 1 2 1.748504e01 0.000e+00 1.734e01 2.260e01 2 4 1.570560e01 0.000e+00 2.608e01 9.347e01 3 6 6.629161e02 0.000e+00 1.241e01 3.103e01 4 7 1.584082e01 0.000e+00 7.934e02 1.826e01 5 8 2.349124e01 0.000e+00 1.912e02 1.571e01 6 9 2.255299e01 0.000e+00 1.955e02 1.993e02 7 10 2.444225e01 0.000e+00 4.293e03 3.821e02 8 11 2.446931e01 0.000e+00 8.100e04 4.035e03 9 12 2.446933e01 0.000e+00 1.999e04 8.126e04 10 13 2.448531e01 0.000e+00 4.004e05 3.289e04 11 14 2.448927e01 0.000e+00 4.036e07 8.156e05 12 15 2.448931e01 0.000e+00 4.000e09 8.230e07 Local minimum found that satisfies the constraints. Optimization completed because the objective function is nondecreasing in feasible directions, to within the selected value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
We now choose to see more decimals in the solution, in order to see more accurately the difference that the new tolerances make.
format long
The optimizer found a solution at:
x
x = 0.972742227363546 0.468569289098342
Compare this to the previous value:
xold
xold = 0.972742694488360 0.468569966693330
The change is
x  xold
ans = 1.0e06 * 0.467124813385844 0.677594988729435
The function value at the solution is:
fval
fval = 0.244893137879894
The solution improved by
fval  minfval
ans = 3.996450220755676e07
(this is negative since the new solution is smaller)
The total number of function evaluations was:
output.funcCount
ans = 15
Compare this to the number of function evaluations when the problem is solved with userprovided gradients but with the default tolerances:
Fgradevals
Fgradevals = 14
Constrained Optimization Example with UserSupplied Hessian
If you give not only a gradient, but also a Hessian, solvers are even more accurate and efficient.
fmincon's interiorpoint solver takes a Hessian matrix as a separate function (not part of the objective function). The Hessian function H(x,lambda) should evaluate the Hessian of the Lagrangian; see the User's Guide for the definition of this term.
Solvers calculate the values lambda.ineqnonlin and lambda.eqlin; your Hessian function tells solvers how to use these values.
In this problem we have but one inequality constraint, so the Hessian is:
type hessfordemo
function H = hessfordemo(x,lambda) % HESSFORDEMO Helper function for Tutorial for the Optimization Toolbox demo % Copyright 20082009 The MathWorks, Inc. s = exp((x(1)^2+x(2)^2)); H = [2*x(1)*(2*x(1)^23)*s+1/10, 2*x(2)*(2*x(1)^21)*s; 2*x(2)*(2*x(1)^21)*s, 2*x(1)*(2*x(2)^21)*s+1/10]; hessc = [2,1/2;1/2,1]; H = H + lambda.ineqnonlin(1)*hessc;
In order to use the Hessian, you need to set options appropriately:
options = optimoptions('fmincon','Algorithm','interiorpoint','GradConstr','on',... 'GradObj','on','Hessian','usersupplied','HessFcn',@hessfordemo);
The tolerances have been set back to the defaults. There should be fewer function counts this time.
options.Display = 'iter';
Call the solver:
[x,fval,exitflag,output] = fmincon(@onehump,x0,[],[],[],[],[],[], ...
@tiltellipse,options);
Firstorder Norm of Iter Fcount f(x) Feasibility optimality step 0 1 2.365241e01 0.000e+00 1.972e01 1 3 5.821325e02 0.000e+00 1.443e01 8.728e01 2 5 1.218829e01 0.000e+00 1.007e01 4.927e01 3 6 1.421167e01 0.000e+00 8.486e02 5.165e02 4 7 2.261916e01 0.000e+00 1.989e02 1.667e01 5 8 2.433609e01 0.000e+00 1.537e03 3.486e02 6 9 2.446875e01 0.000e+00 2.057e04 2.727e03 7 10 2.448911e01 0.000e+00 2.068e06 4.191e04 8 11 2.448931e01 0.000e+00 2.001e08 4.218e06 Local minimum found that satisfies the constraints. Optimization completed because the objective function is nondecreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
There were fewer, and different iterations this time.
The solution to this problem has been found at:
x
x = 0.972742246093537 0.468569316215571
The function value at the solution is:
fval
fval = 0.244893121872758
The total number of function evaluations was:
output.funcCount
ans = 11
Compare this to the number using only gradient evaluations, with the same default tolerances:
Fgradevals
Fgradevals = 14