optimization expression includes an integration

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Sukvasant Tantikovt
编辑： Torsten 2024-7-29
I am trying an optimization problem in which the expression of the objective function includes an integral.
It is obvious that the \sigma equal to one results in the optimium solution. I want to use the optimization toolbox to get this result with an initial \sigma equal to, say, 10.
I wrote the following code.
g1 = @(x,c) (exp(-(0.5*(x./c).^2))./sqrt(2*pi*c^2));
c = optimvar("c",1,1,'Type','continuous','LowerBound',0.1,'UpperBound',10);
prob = optimproblem('Objective', (0.5 - integral(@(x)g1(x,c),0, 10)).^2);
[solf,fvalf,eflagf,outputf] = solve(prob)
The following error is generated.
Error using integralCalc>finalInputChecks (line 544)
Input function must return 'double' or 'single' values. Found
'optim.problemdef.OptimizationExpression'.
I have two questions:
1, Am I coding the problem properly/correctly?
2, If the code is basically correct, how can I solve the error?
Thank you.

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采纳的回答

Torsten
Torsten 2024-7-25
编辑：Torsten 2024-7-25
Ran in:
c = optimvar("c",1,1,'Type','continuous','LowerBound',0.1,'UpperBound',10);
g1 = @(x,c) exp(-0.5*(x./c).^2)./sqrt(2*pi*c^2);
obj = fcn2optimexpr(@(c)abs(0.5-integral(@(x)g1(x,c),0,10)).^0.5,c);
prob = optimproblem('Objective', obj);
show(prob)
OptimizationProblem : Solve for: c minimize : arg1 where: anonymousFunction1 = @(c)abs(0.5-integral(@(x)g1(x,c),0,10)).^0.5; arg1 = anonymousFunction1(c); variable bounds: 0.1 <= c <= 10
x0.c = 10;
options = optimoptions("fmincon",OptimalityTolerance=1e-18);
[solf,fvalf,eflagf,outputf] = solve(prob,x0,Options=options)
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.
solf = struct with fields:
c: 1.0115
fvalf = 0
eflagf =
OptimalSolution
outputf = struct with fields:
iterations: 6 funcCount: 35 constrviolation: 0 stepsize: 6.9192e-06 algorithm: 'interior-point' firstorderopt: 0 cgiterations: 18 message: 'Local minimum found that satisfies the constraints....' bestfeasible: [1x1 struct] objectivederivative: "finite-differences" constraintderivative: "closed-form" solver: 'fmincon'
%The function showing the error is very flat - thus c = 1 is unlikely as
%result.
y = 0.01:0.01:3;
G1 = arrayfun(@(y)abs(0.5-integral(@(x)g1(x,y),0,10)).^0.5,y);
plot(y,G1)
grid on
  2 个评论
Sukvasant Tantikovt
Sukvasant Tantikovt 2024-7-29
Hi Torsten,
Thank you for your help.
The solution seems to be very sensitivty to the upper bound value. If this value is changed to 30, the code produces c equal to 1.5 747. If the upper bound is 20, c = 0.4068 is obtained. Do you have any idea?
Torsten
Torsten 2024-7-29
编辑：Torsten 2024-7-29
I plotted the error function for your problem. It's almost 0 for c in the interval [0 3] and too flat to get good convergence. Most probably, the integral cannot be computed with sufficient accuracy. Using MATLAB's "erf" function (after a coordinate transformation) or a symbolic computation might help.

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