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Objective and Constraints Having a Common Function in Serial or Parallel, Problem-Based

This example shows how to avoid calling a function twice when it computes values for both the objective and the constraints using the problem-based approach. For the solver-based approach, see Objective and Nonlinear Constraints in the Same Function.

You typically use such a function in a simulation. Solvers usually evaluate the objective and nonlinear constraint functions separately. This evaluation is wasteful when you use the same calculation for both results.

This example also shows the effect of parallel computation on solver speed. For time-consuming functions, computing in parallel can speed the solver, as can avoiding calling the time-consuming function repeatedly at the same point. Using both techniques together speeds the solver the most.

Create Time-Consuming Function That Computes Several Quantities

The computeall function returns outputs that are part of the objective and nonlinear constraints.

type computeall
function [f1,c1] = computeall(x)
    c1 = norm(x)^2 - 1;
    f1 = 100*(x(2) - x(1)^2)^2 + (1 - x(1))^2;
    pause(1) % simulate expensive computation
end

The function includes a pause(1) statement to simulate a time-consuming function.

Create Optimization Variables

This problem uses a four-element optimization variable.

x = optimvar('x',4);

Convert Function Using 'ReuseEvaluation'

Convert the computeall function to an optimization expression. To save time during the optimization, use the 'ReuseEvaluation' name-value pair. To save time for the solver to determine the output expression sizes (this happens only once), set the 'OutputSize' name-value pair to [1 1], indicating that both f and c are scalar.

[f,c] = fcn2optimexpr(@computeall,x,'ReuseEvaluation',true,'OutputSize',[1 1]);

Create Objective, Constraint, and Problem

Create the objective function from the f expression.

obj = f + 20*(x(3) - x(4)^2)^2 + 5*(1 - x(4))^2;

Create the nonlinear inequality constraint from the c expression.

cons = c <= 0;

Create an optimization problem and include the objective and constraint.

prob = optimproblem('Objective',obj);
prob.Constraints.cons = cons;
showproblem(prob)
  OptimizationProblem : 

	minimize :
  ((arg3 + (20 .* (x(3) - x(4).^2).^2)) + (5 .* (1 - x(4)).^2))

  where:

    [arg3,~] = generatedFunction_computeall_withReuse(x);

	subject to cons:
  arg_LHS <= 0

  where:

    [~,arg_LHS] = generatedFunction_computeall_withReuse(x);

Solve Problem

Monitor the time it takes to solve the problem starting from the initial point x0.x = [-1;1;1;2].

x0.x = [-1;1;1;2];
x0.x = x0.x/norm(x0.x); % Feasible initial point
tic
[sol,fval,exitflag,output] = solve(prob,x0)
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: [4×1 double]

fval = 0.7107
exitflag = 
    OptimalSolution

output = struct with fields:
         iterations: 25
          funcCount: 149
    constrviolation: 0
           stepsize: 1.3715e-07
          algorithm: 'interior-point'
      firstorderopt: 4.0000e-07
       cgiterations: 7
            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, 2.909695e-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'

time1 = toc
time1 = 149.8888

The number of seconds for the solution is just over the number of function evaluations, which indicates that the solver computed each evaluation only once.

fprintf("The number of seconds to solve was %g, and the number of evaluation points was %g.\n",time1,output.funcCount)
The number of seconds to solve was 149.889, and the number of evaluation points was 149.

If, instead, you do not call fcn2optimexpr using 'ReuseEvaluation', then the solution time doubles.

[f2,c2] = fcn2optimexpr(@computeall,x,'ReuseEvaluation',false);
obj2 = f2 + 20*(x(3) - x(4)^2)^2 + 5*(1 - x(4))^2;
cons2 = c2 <= 0;
prob2 = optimproblem('Objective',obj2);
prob2.Constraints.cons2 = cons2;
tic
[sol2,fval2,exitflag2,output2] = solve(prob2,x0);
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>
time2 = toc
time2 = 298.4074

Parallel Processing

If you have a Parallel Processing Toolbox™ license, you can save even more time by computing in parallel. To do so, set options to use parallel processing, and call solve with options.

options = optimoptions(prob,'UseParallel',true);
tic
[sol3,fval3,exitflag3,output3] = solve(prob,x0,'Options',options);
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>
time3 = toc
time3 = 74.0615

Using parallel processing and 'ReuseEvaluation' together provides a faster solution than using 'ReuseEvaluation' alone. See how long it takes to solve the problem using parallel processing alone.

tic
[sol4,fval4,exitflag4,output4] = solve(prob2,x0,'Options',options);
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>
time4 = toc
time4 = 144.6696

Summary of Timing Results

Combine the timing results into one table.

timingtable = table([time1;time2;time3;time4],...
    'RowNames',["Reuse Serial";"No Reuse Serial";"Reuse Parallel";"No Reuse Parallel"])
timingtable=4×1 table
                          Var1 
                         ______

    Reuse Serial         149.89
    No Reuse Serial      298.41
    Reuse Parallel       74.062
    No Reuse Parallel    144.67

For this problem, on a computer with a 6-core processor, computing in parallel takes about half the time of computing in serial, and computing with 'ReuseEvaluation' takes about half the time of computing without 'ReuseEvaluation'. Computing in parallel with 'ReuseEvaluation' takes about a quarter of the time of computing in serial without 'ReuseEvaluation'.

See Also

Related Topics