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Solve Constrained Nonlinear Optimization, Problem-Based

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This example shows how to find the minimum of a nonlinear objective function with a nonlinear constraint by using the problem-based approach. For a video showing the solution to a similar problem, seeProblem-Based Nonlinear Programming.

To find the minimum value of a nonlinear objective function using the problem-based approach, first write the objective function as a file or anonymous function. The objective function for this example is

f ( x , y ) = e x ( 4 x 2 + 2 y 2 + 4 x y + 2 y - 1 ) . 

typeobjfunx
function f = objfunx(x,y) f = exp(x).*(4*x.^2 + 2*y.^2 + 4*x.*y + 2*y - 1); end

Create the optimization problem variablesxandy.

x = optimvar('x');y = optimvar('y');

Create the objective function as an expression in the optimization variables.

obj = objfunx(x,y);

Create an optimization problem withobjas the objective function.

prob = optimproblem('Objective',obj);

Create a nonlinear constraint that the solution lies in a tilted ellipse, specified as

x y 2 + ( x + 2 ) 2 + ( y - 2 ) 2 2 ≤ 2 .

Create the constraint as an inequality expression in the optimization variables.

TiltEllipse = x.*y/2 + (x+2).^2 + (y-2).^2/2 <= 2;

Include the constraint in the problem.

prob.Constraints.constr = TiltEllipse;

Create a structure representing the initial point asx = –3,y = 3.

x0。x = 3;x0。y = 3;

Review the problem.

show(prob)
OptimizationProblem : Solve for: x, y minimize : (exp(x) .* (((((4 .* x.^2) + (2 .* y.^2)) + ((4 .* x) .* y)) + (2 .* y)) - 1)) subject to constr: ((((x .* y) ./ 2) + (x + 2).^2) + ((y - 2).^2 ./ 2)) <= 2

Solve the problem.

[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: -5.2813 y: 4.6815
fval = 0.3299

Try a different start point.

x0。x = -1; x0.y = 1; [sol2,fval2] = 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.
sol2 =struct with fields:x: -0.8210 y: 0.6696
fval2 = 0.7626

Plot the ellipse, the objective function contours, and the two solutions.

f = @objfunx; g = @(x,y) x.*y/2+(x+2).^2+(y-2).^2/2-2; rnge = [-5.5 -0.25 -0.25 7]; fimplicit(g,'k-') axis(rnge); holdonfcontour(f,rnge,'LevelList',logspace(-1,1)) plot(sol.x,sol.y,'ro','LineWidth',2) plot(sol2.x,sol2.y,'ko','LineWidth',2) legend('Constraint','f Contours','Global Solution','Local Solution','Location','northeast');持有的f

Figure contains an axes object. The axes object contains 4 objects of type implicitfunctionline, functioncontour, line. These objects represent Constraint, f Contours, Global Solution, Local Solution.

The solutions are on the nonlinear constraint boundary. The contour plot shows that these are the only local minima. The plot also shows that there is a stationary point near [–2,3/2], and local maxima near [–2,0] and [–1,4].

Convert Objective Function Usingfcn2optimexpr

For some objective functions or software versions, you must convert nonlinear functions to optimization expressions by usingfcn2optimexpr. See金宝app支持操作一个优化变量d ExpressionsandConvert Nonlinear Function to Optimization Expression. Pass thexandyvariables in thefcn2optimexprcall to indicate which optimization variable corresponds to eachobjfunxinput.

obj = fcn2optimexpr(@objfunx,x,y);

Create an optimization problem withobjas the objective function just as before.

prob = optimproblem('Objective',obj);

The remainder of the solution process is identical.

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