To use a nonlinear function as an objective or nonlinear constraint function in the problem-based approach, convert the function to an optimization expression using `fcn2optimexpr`

. This example shows how to convert the function using both a function file and an anonymous function.

To use a function file in the problem-based approach, you need to convert the file to an expression using `fcn2optimexpr`

.

For example, the `expfn3.m`

file contains the following code:

`type expfn3.m`

function [f,g,mineval] = expfn3(u,v) mineval = min(eig(u)); f = v'*u*v; f = -exp(-f); t = u*v; g = t'*t + sum(t) - 3;

To use this function file as an optimization expression, first create optimization variables of the appropriate sizes.

u = optimvar('u',3,3,'LowerBound',-1,'UpperBound',1); % 3-by-3 variable v = optimvar('v',3,'LowerBound',-2,'UpperBound',2); % 3-by-1 variable

Convert the function file to an optimization expressions using `fcn2optimexpr`

.

[f,g,mineval] = fcn2optimexpr(@expfn3,u,v);

Because all returned expressions are scalar, you can save computing time by specifying the expression sizes using the `'OutputSize'`

name-value pair argument. Also, because `expfn3`

computes all of the outputs, you can save more computing time by using the `ReuseEvaluation`

name-value pair.

[f,g,mineval] = fcn2optimexpr(@expfn3,u,v,'OutputSize',[1,1],'ReuseEvaluation',true)

f = Nonlinear OptimizationExpression [argout,~,~] = expfn3(u, v)

g = Nonlinear OptimizationExpression [~,argout,~] = expfn3(u, v)

mineval = Nonlinear OptimizationExpression [~,~,argout] = expfn3(u, v)

To use a general nonlinear function handle in the problem-based approach, convert the handle to an optimization expression using `fcn2optimexpr`

. For example, write a function handle equivalent to `f`

and convert it.

```
fun = @(x,y)-exp(-y'*x*y);
funexpr = fcn2optimexpr(fun,u,v,'OutputSize',[1,1])
```

funexpr = Nonlinear OptimizationExpression anonymousFunction1(u, v) where: anonymousFunction1 = @(x,y)-exp(-y'*x*y);

To use either expression as an objective function, create an optimization problem.

```
prob = optimproblem;
prob.Objective = f;
% Or, equivalently, prob.Objective = funexpr;
```

Define the constraint `g <= 0`

in the optimization problem.

prob.Constraints.nlcons1 = g <= 0;

Also define the constraints that `u`

is symmetric and that $$mineval\ge -1/2$$.

prob.Constraints.sym = u == u.'; prob.Constraints.mineval = mineval >= -1/2;

View the problem.

show(prob)

OptimizationProblem : Solve for: u, v minimize : [argout,~,~] = expfn3(u, v) subject to nlcons1: arg_LHS <= 0 where: [~,arg_LHS,~] = expfn3(u, v); subject to sym: u(2, 1) - u(1, 2) == 0 u(3, 1) - u(1, 3) == 0 -u(2, 1) + u(1, 2) == 0 u(3, 2) - u(2, 3) == 0 -u(3, 1) + u(1, 3) == 0 -u(3, 2) + u(2, 3) == 0 subject to mineval: arg_LHS >= (-0.5) where: [~,~,arg_LHS] = expfn3(u, v); variable bounds: -1 <= u(1, 1) <= 1 -1 <= u(2, 1) <= 1 -1 <= u(3, 1) <= 1 -1 <= u(1, 2) <= 1 -1 <= u(2, 2) <= 1 -1 <= u(3, 2) <= 1 -1 <= u(1, 3) <= 1 -1 <= u(2, 3) <= 1 -1 <= u(3, 3) <= 1 -2 <= v(1) <= 2 -2 <= v(2) <= 2 -2 <= v(3) <= 2

To solve the problem, call `solve`

. Set an initial point `x0`

.

rng default % For reproducibility x0.u = 0.25*randn(3); x0.u = x0.u + x0.u.'; x0.v = 2*randn(3,1); [sol,fval,exitflag,output] = solve(prob,x0)

Solving problem using fmincon. Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

`sol = `*struct with fields:*
u: [3x3 double]
v: [3x1 double]

fval = -403.4288

exitflag = OptimalSolution

`output = `*struct with fields:*
iterations: 87
funcCount: 1448
constrviolation: 6.3860e-12
stepsize: 7.4093e-05
algorithm: 'interior-point'
firstorderopt: 0.0012
cgiterations: 172
message: '...'
solver: 'fmincon'

View the solution.

disp(sol.u)

0.8419 0.5748 -0.7670 0.5748 0.3745 0.2997 -0.7670 0.2997 0.5667

disp(sol.v)

2.0000 -2.0000 2.0000

The solution matrix `u`

is symmetric. All values of `v`

are at the bounds.