Initialize Optimization Expressions - MATLAB & Simulink (original) (raw)

This topic describes an error that occurs when an optimization expression is not properly initialized, and provides examples of different ways to address the error.

Error in Expression

Sometimes you receive this error from an objective or nonlinear constraint function or expression:

Unable to perform assignment because value of type 'optim.problemdef.OptimizationExpression' is not convertible to 'double'.

Often, this error results from an improper initialization of an optimization expression. Typically, you initialize a variable F in a standard loop by declaring an array of zeros, such as:

However, if F is an optimization expression, then you must initialize it using optimexpr.

This topic provides examples of proper initialization techniques. All are based on the same example, a function that uses an internal loop.

function f = myFun(x) out = zeros(size(x)); out(1) = x(1); for i = 2:10 out(i) = (x(i) - x(i-1))^3; end f = mean(out); end

If you try to use myFun(x) as the objective function for an optimization variable x, you get an error.

x = optimvar("x",10,LowerBound=0,UpperBound=10); prob = optimproblem(Objective=myFun(x));

Unable to perform assignment because value of type 'optim.problemdef.OptimizationVariable' is not convertible to 'double'.

Error in myFun (line 3) out(1) = x(1);

Caused by: Error using double Conversion to double from optim.problemdef.OptimizationVariable is not possible.

However, myFun works as the objective in a solver-based problem.

rng default x0 = 10*rand(10,1); lb = zeros(10,1); ub = 10 + lb; [sol,fval] = fmincon(@myFun,x0,[],[],[],[],lb,ub)

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 =

9.4226

10.0000 0.0000 5.0000 10.0000 0.0000 3.3333 6.6667 10.0000 0.0000

fval =

-262.9274

This problem has several local solutions, so you might get different answers depending on your initial point.

Use "like" Syntax to Initialize the Array

Rewrite the function to use the "like" argument forzeros. You can then pass an optimization expression or a numeric array as input.

This approach has several advantages:

• The function signature is maintained.
• The approach does not introduce extra code that runs during the solution process.
• If applicable, automatic differentiation is enabled.
• You can use the function in either the solver-based or problem-based workflow.
• The approach enables static analysis when usingfcn2optimexpr. For information about static analysis, see Static Analysis of Optimization Expressions.

function f = myFun1(x) out = zeros(size(x),"like",x); out(1) = x(1); for i = 2:10 out(i) = (x(i) - x(i-1))^3; end f = mean(out); end

Use myFun1 in a problem-based workflow.

x = optimvar("x",10,LowerBound=0,UpperBound=10); prob = optimproblem(Objective=myFun1(x)); rng default x0.x = 10*rand(10,1); [sol,fval] = 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:

x: [10×1 double]

fval =

-262.9274

Use myFun1 in a solver-based workflow.

rng default x0 = 10*rand(10,1); lb = zeros(10,1); ub = 10 + lb; [sol,fval] = fmincon(@myFun1,x0,[],[],[],[],lb,ub)

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 =

9.4226

10.0000 0.0000 5.0000 10.0000 0.0000 3.3333 6.6667 10.0000 0.0000

fval =

-262.9274

Modify Function to Accept an Initial Array

Rewrite the function to accept the initial value as an additional argument. You can then pass an optimization expression or a numeric array as the initial value.myFun2 uses the input variable out as the output variable, and accepts either a zero array or an optimization expression.

This approach has several advantages:

• If applicable, automatic differentiation is enabled.
• The approach does not introduce extra code that runs during the solution process.
• You can use the function in either the solver-based or problem-based workflow.
• The approach enables static analysis when usingfcn2optimexpr. For information about static analysis, see Static Analysis of Optimization Expressions.

The disadvantage of this approach is that you must rewrite the function with a different function signature.

function f = myFun2(out,x) out(1) = x(1); for i = 2:10 out(i) = (x(i) - x(i-1))^3; end f = mean(out); end

Use myFun2 in a problem-based workflow.

x = optimvar("x",10,LowerBound=0,UpperBound=10); out = optimexpr(size(x)); prob = optimproblem(Objective=myFun2(out,x)); rng default x0.x = 10*rand(10,1); [sol,fval] = 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:

x: [10×1 double]

fval =

-262.9274

Use myFun2 in a solver-based workflow.

rng default x0 = 10*rand(10,1); lb = zeros(10,1); ub = 10 + lb; out = zeros(size(x0)); [sol,fval] = fmincon(@(x)myFun2(out,x),x0,[],[],[],[],lb,ub)

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 =

9.4226

10.0000 0.0000 5.0000 10.0000 0.0000 3.3333 6.6667 10.0000 0.0000

fval =

-262.9274

Rewrite Function to Initialize Expressions Appropriately

You can explicitly check for the type of problem variables and initialize an expression appropriately. This method has several advantages:

• If applicable, automatic differentiation is enabled.
• You can use the function in either the solver-based or problem-based workflow.

The disadvantages of this approach are that you must rewrite the function, a small amount of overhead is introduced while the solver runs, and, because of theif statement, static analysis is not supported when usingfcn2optimexpr. For information about static analysis, seeStatic Analysis of Optimization Expressions.

function f = myFun3(x) % Check for the data type of variable x if isa(x,'double') out = zeros(size(x)); else out = optimexpr(size(x)); end % No changes to the rest of the code out(1) = x(1); for i = 2:10 out(i) = (x(i) - x(i-1))^3; end f = mean(out); end

Solve the problem using optimization variables with the objective functionmyFun3.

x = optimvar("x",10,LowerBound=0,UpperBound=10); prob = optimproblem(Objective=myFun3(x)); rng default x0.x = 10*rand(10,1); [sol,fval] = 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:

x: [10×1 double]

fval =

-262.9274

Solve the problem using fmincon with the objective functionmyFun3.

rng default x0 = 10*rand(10,1); lb = zeros(10,1); ub = 10 + lb; out = zeros(size(x0)); [sol,fval] = fmincon(@myFun3,x0,[],[],[],[],lb,ub)

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 =

9.4226

10.0000 0.0000 5.0000 10.0000 0.0000 3.3333 6.6667 10.0000 0.0000

fval =

-262.9274

Use fcn2optimexpr Conversion

For R2022b and later, you can convert the objective function to an optimization expression using fcn2optimexpr, and then initialize the expression using a standard zero array.

This approach has several advantages:

• The function signature is maintained.
• If applicable, automatic differentiation is enabled.
• You can use the function in either the solver-based or problem-based workflow.

This approach requires static analysis, so it might not run correctly in earlier MATLAB® versions, and might incur a bit of overhead. This example uses the original function myFun, which fails in the problem-based workflow in Error in Expression.

x = optimvar("x",10,LowerBound=0,UpperBound=10); obj = fcn2optimexpr(@myFun,x,Display="on"); prob = optimproblem(Objective=obj); rng default x0.x = 10*rand(10,1); [sol,fval] = solve(prob,x0)

The function uses only supported operators. The returned expressions use the operators on the problem variables.

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:

x: [10×1 double]

fval =

-262.9274

See Also

fcn2optimexpr | optimexpr

Topics

• Problem-Based Optimization Workflow
• Static Analysis of Optimization Expressions