fcn2optimexpr

Convert function to optimization expression

Description

example

[out1,out2,...,outN] = fcn2optimexpr(fcn,in1,in2,...,inK) converts the function fcn(in1,in2,...,inK) to an optimization expression having N outputs.

example

[out1,out2,...,outN] = fcn2optimexpr(fcn,in1,in2,...,inK,Name,Value) modifies the expression creation process according to name-value parameters.

Examples

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To use a MATLAB™ function in the problem-based approach, first convert it to an optimization expression.

For example, to use the objective function -exp(-x2/2), create an optimization variable x and use it in a converted anonymous function:

x = optimvar('x');
obj = fcn2optimexpr(@(t)-exp(-t^2/2),x);
prob = optimproblem('Objective',obj);
showproblem(prob)
  OptimizationProblem : 

	Solve for:
       x

	minimize :
       anonymousFunction1(x)

       where:

         anonymousFunction1 = @(t)-exp(-t^2/2);

For more complex functions, convert a function file. For example, suppose that you have a function file named expfn2.m that computes an objective in two optimization variables.

type expfn2
function f = expfn2(t,u)
f = -exp(-t^2/2)*u/(1 + u^2);

Include this objective in a problem.

x = optimvar('x');
y = optimvar('y','LowerBound',0);
obj = fcn2optimexpr(@expfn2,x,y);
prob = optimproblem('Objective',obj);
showproblem(prob)
  OptimizationProblem : 

	Solve for:
       x, y

	minimize :
       expfn2(x, y)


	variable bounds:
       0 <= y

If your function has several outputs, you can use them as elements of the objective function. For example, suppose that u is a 2-by-2 variable and v is a 2-by-1 variable, and expfn3 has three outputs:

type expfn3
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;

Create appropriately sized optimization variables, and create an objective function from the first two outputs.

u = optimvar('u',2,2);
v = optimvar('v',2);
[f,g,mineval] = fcn2optimexpr(@expfn3,u,v);
prob = optimproblem;
prob.Objective = f*g/(1 + f^2);
showproblem(prob)
  OptimizationProblem : 

	Solve for:
       u, v

	minimize :
       ((arg2 .* arg3) ./ (1 + arg1.^2))

       where:

         [arg1,~,~] = expfn3(u, v);
         [arg2,~,~] = expfn3(u, v);
         [~,arg3,~] = expfn3(u, v);

You can use the mineval output in a subsequent constraint expression.

In problem-based optimization, constraints are two optimization expressions with a comparison operator (==, <=, or >=) between them. You can use fcn2optimexpr to create one or both optimization expressions.

Create the constraint that expfn2 is less than or equal to –1/2. This function of two variables is in the expfn2.m file.

type expfn2
function f = expfn2(t,u)
f = -exp(-t^2/2)*u/(1 + u^2);

Create optimization variables, convert the function file to an optimization expression, then express the constraint as confn.

x = optimvar('x');
y = optimvar('y','LowerBound',0);
expr1 = fcn2optimexpr(@expfn2,x,y);
confn = expr1 <= -1/2;
showconstr(confn)
  expfn2(x, y) <= -0.5

Create another constraint that expfn2 is greater than x + y.

confn2 = expr1 >= x + y;

Create an optimization problem and place the constraints in the problem.

prob = optimproblem;
prob.Constraints.confn = confn;
prob.Constraints.confn2 = confn2;
showproblem(prob)
  OptimizationProblem : 

	Solve for:
       x, y

	minimize :

	subject to confn:
       expfn2(x, y) <= -0.5

	subject to confn2:
       expfn2(x, y) >= (x + y)

	variable bounds:
       0 <= y

If the objective and nonlinear constraint come from a common, time-consuming function, save time by using the 'ReuseEvaluation' name-value pair. For example, rosenbrocknorm computes both the Rosenbrock objective function, and the norm of the argument for use in the constraint x24.

type rosenbrocknorm
function [f,c] = rosenbrocknorm(x)
pause(1) % Simulates time-consuming function
c = dot(x,x);
f = 100*(x(2) - x(1)^2)^2 + (1 - x(1))^2;

Create a 2-D optimization variable x. Then convert rosenbrocknorm to an optimization expression by using fcn2optimexpr, specifying 'ReuseEvaluation'.

x = optimvar('x',2);
[f,c] = fcn2optimexpr(@rosenbrocknorm,x,'ReuseEvaluation',true);

Create objective and constraint expressions from the returned expressions. Include these expressions in an optimization problem. Review the problem using showproblem.

prob = optimproblem('Objective',f);
prob.Constraints.cineq = c <= 4;
showproblem(prob)
  OptimizationProblem : 

	Solve for:
       x

	minimize :
       [argout,~] = rosenbrocknorm(x)


	subject to cineq:
       arg_LHS <= 4

       where:

         [~,arg_LHS] = rosenbrocknorm(x);
     

Solve the problem starting from the initial point x0.x = [-1;1], timing the result.

x0.x = [-1;1];
tic
[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.

<stopping criteria details>
sol = struct with fields:
    x: [2×1 double]

fval = 3.6222e-11
exitflag = 
    OptimalSolution

output = struct with fields:
         iterations: 43
          funcCount: 161
    constrviolation: 0
           stepsize: 9.1067e-08
          algorithm: 'interior-point'
      firstorderopt: 6.3912e-07
       cgiterations: 10
            message: '↵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.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 6.391223e-07,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 0.000000e+00, is less than options.ConstraintTolerance = 1.000000e-06.↵↵'
             solver: 'fmincon'

toc
Elapsed time is 161.656507 seconds.

The solution time in seconds is nearly the same as the number of function evaluations. This result indicates that the solver reused function values, and did not waste time by reevaluating the same point twice.

For a more extensive example, see Objective and Constraints Having a Common Function in Serial or Parallel, Problem-Based.

Input Arguments

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Function to convert, specified as a function handle.

Example: @sin specifies the sine function

Data Types: function_handle

Input argument, specified as a MATLAB variable. The input can have any data type and any size.

Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | char | string | struct | table | cell | function_handle | categorical | datetime | duration | calendarDuration | fi
Complex Number Support: Yes

Name-Value Pair Arguments

Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.

Example: [out1,out2] = fcn2optimexpr(@fun,x,y,'OutputSize',[1,1],'Reuse',true) specifies that out1 and out2 are scalars and that these variables will be reused between objective and constraint functions without recalculation.

Sizes of output expressions, specified as:

  • An integer vector — If there is one output out1, OutputSize specifies the size of out1. If there are multiple outputs out1,…,outN, OutputSize specifies that all outputs have the same size.

    Note

    A scalar has size [1,1].

  • A cell array of integer vectors — The size of output outj is the jth element of OutputSize.

If you do not pass an OutputSize name-value pair, then fcn2optimexpr passes data to fcn in order to determine the sizes of the outputs (see Algorithms). By passing OutputSize, you enable fcn2optimexpr to skip this step. Skipping this evaluation saves time. Additionally, if you do not pass an OutputSize name-value pair, and if the evaluation of fcn fails for any reason, then fcn2optimexpr fails as well.

Example: [out1,out2,out3] = fcn2optimexpr(@fun,x,'OutputSize',[1,1]) specifies that the three outputs [out1,out2,out3] are scalars

Example: [out1,out2] = fcn2optimexpr(@fun,x,'OutputSize',{[4,4],[3,5]}) specifies that out1 has size 4-by-4 and out2 has size 3-by-5.

Data Types: double | cell

Enable reusable values, specified as false (do not enable) or true (enable).

ReuseEvaluation can make your problem run faster when, for example, the objective and some nonlinear constraints rely on a common calculation. In this case, the solver stores the value for reuse wherever needed, so avoids recalculating the value.

There is a small overhead in enabling reusable values, so it is best to enable reusable values only for expressions that share a value.

Example: [out1,out2,out3] = fcn2optimexpr(@fun,x,'ReuseEvaluation',true) allows out1, out2, and out3 to be used in multiple computations, with the outputs being calculated only once per evaluation point

Data Types: logical

Output Arguments

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Output argument, returned as an OptimizationExpression. The size of the expression depends on the input function.

Algorithms

To find the output size of each returned expression when you do not provide an OutputSize, fcn2optimexpr evaluates your function at the following point for each element of the problem variables.

  • If there is a finite upper bound ub and a finite lower bound lb, the evaluation point is (lb + ub)/2 + ((ub - lb)/2)*eps.

  • If there is a finite lower bound and no upper bound, the evaluation point is lb + max(1,abs(lb))*eps.

  • If there is a finite upper bound and no lower bound, the evaluation point is ub - max(1,abs(ub))*eps.

  • If there are no bounds, the evaluation point is 1 + eps.

  • In addition, if the variable is specified as an integer, the evaluation point is floor of the point given previously.

It is possible that this evaluation point leads to an error in function evaluation. To avoid this error, specify OutputSize.

Introduced in R2019a