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.mfunction [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 .
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: 74
funcCount: 1156
constrviolation: 0
stepsize: 1.6235e-06
algorithm: 'interior-point'
firstorderopt: 1.1203e-04
cgiterations: 122
message: '...'
solver: 'fmincon'
View the solution.
disp(sol.u)
0.5486 0.2067 -0.8420
0.2067 0.1909 0.4842
-0.8420 0.4842 0.8262
disp(sol.v)
2.0000
-2.0000
2.0000
The solution matrix u is symmetric. All values of v are at the bounds.