This section shows how to choose whether to convert a nonlinear function to an optimization expression or to create the expression out of supported operations on optimization variables. The section also shows how to convert a function, if necessary, by using `fcn2optimexpr`

.

Generally, create your objective or nonlinear constraint functions by using supported operations on optimization variables and expressions. Doing so has these advantages:

`solve`

includes gradients calculated by automatic differentiation. See Effect of Automatic Differentiation in Problem-Based Optimization.`solve`

has a wider choice of available solvers. When using`fcn2optimexpr`

,`solve`

uses only`fmincon`

or`fminunc`

.

In general, supported operations include all elementary mathematical operations: addition, subtraction, multiplication, division, powers, and elementary functions such as exponential and trigonometric functions and their inverses. Nonsmooth operations such as `max`

, `abs`

, `if`

, and `case`

are not supported. For the complete description, see Supported Operations on Optimization Variables and Expressions.

For example, suppose that your objective function is

$$f(x,y,r)=100(y-{x}^{2}{)}^{2}+(r-x{)}^{2}$$

where $$r$$ is a parameter that you supply, and the problem is to minimize $$f$$ over $$x$$ and $$y$$ . This objective function is a sum of squares, and takes the minimal value of 0 at the point $$x=r$$, $$y={r}^{2}$$.

The objective function is a polynomial, so you can write it in terms of elementary operations on optimization variables.

r = 2; x = optimvar('x'); y = optimvar('y'); f = 100*(y - x^2)^2 + (r - x)^2; prob = optimproblem("Objective",f); x0.x = -1; x0.y = 2; [sol,fval] = solve(prob,x0)

Solving problem using lsqnonlin. Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.

`sol = `*struct with fields:*
x: 2.0000
y: 4.0000

fval = 1.4644e-20

To solve the same problem by converting the objective function using `fcn2optimexpr`

(not recommended), first write the objective as an anonymous function.

fun = @(x,y)100*(y - x^2)^2 + (r - x)^2; prob.Objective = fcn2optimexpr(fun,x,y); [sol2,fval2] = solve(prob,x0)

Solving problem using fminunc. Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.

`sol2 = `*struct with fields:*
x: 2.0000
y: 3.9998

fval2 = 1.7143e-09

Notice that `solve`

uses `fminunc`

this time instead of the more efficient `lsqnonlin`

, and the reported solution for `y`

is slightly different than the correct solution 4. Furthermore, the reported `fval`

is about 1e-9 instead of 1e-20 (the actual solution value is exactly 0). These slight inaccuracies are due to `solve`

not using the more efficient solver.

The remainder of this example shows how to convert a function to an optimization expression using `fcn2optimexpr`

.

To use a function file in the problem-based approach, you must 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;

This function is not entirely composed of supported operations because of `min(eig(u))`

. Therefore, to use `expfn3(u,v)`

as an optimization expression, you must first convert it using `fcn2optimexpr`

.

To use `expfn3`

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 `mineval`

and convert it.

```
fun = @(u)min(eig(u));
funexpr = fcn2optimexpr(fun,u,'OutputSize',[1,1])
```

funexpr = Nonlinear OptimizationExpression anonymousFunction2(u) where: anonymousFunction2 = @(u)min(eig(u));

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

prob = optimproblem; prob.Objective = f;

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. Feasible point with lower objective function value found. 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: '...'
bestfeasible: [1x1 struct]
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.

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