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OptimizationValues

Values for optimization problems

Description

An OptimizationValues object holds values used by and returned from solve for multiobjective problems. The object also holds starting values for the x0 argument for solvers that accept multiple start points.

Creation

The solve function returns a vector of OptimizationValues objects as the solution to a multiobjective problem.

Create an OptimizationValues object for a start point x0 by using the optimvalues function.

Properties

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Typically, OptimizationValues properties are dynamic: they are the names of the optimization variables, objective function or functions, and constraints.

However, you can also have unnamed objective functions or constraints. For those cases, OptimizationValues assigns the following properties.

Objective function values, returned or specified as a real array.

Data Types: double

Constraint values, returned or specified as a real array.

Data Types: double

Object Functions

paretoplotPareto plot of multiobjective values

Examples

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Create and solve a multiobjective problem using optimization variables.

x = optimvar("x",LowerBound=-3,UpperBound=3);
prob = optimproblem;
prob.Objective = [x^2;(x-1)^2]; % Tradeoff region between x = 0 and x = 1
prob.Constraints.con1 = x^2 <= 1/2; % Demonstrate constraints
prob.Constraints.con2 = x^2 >= 1/10; % Second constraint
rng default % For reproducibility
[sol,fval,exitflag,output] = solve(prob,Solver="paretosearch")
Solving problem using paretosearch.

Pareto set found that satisfies the constraints. 

Optimization completed because the relative change in the volume of the Pareto set 
is less than 'options.ParetoSetChangeTolerance' and constraints are satisfied to within 
'options.ConstraintTolerance'.
sol = 
  1x60 OptimizationValues vector with properties:

   Variables properties:
            x: [0.6366 0.3952 0.6812 0.5353 0.4299 0.6951 0.3813 0.3223 ... ]

   Objective properties:
    Objective: [2x60 double]

   Constraints properties:
         con1: [-0.0947 -0.3438 -0.0360 -0.2134 -0.3152 -0.0169 -0.3546 ... ]
         con2: [-0.3053 -0.0562 -0.3640 -0.1866 -0.0848 -0.3831 -0.0454 ... ]

fval = 2×60

    0.4053    0.1562    0.4640    0.2866    0.1848    0.4831    0.1454    0.1039    0.1295    0.1016    0.1406    0.3329    0.1300    0.3210    0.2673    0.3223    0.1115    0.4253    0.1170    0.4749    0.3164    0.1099    0.4375    0.1814    0.2936    0.4727    0.4395    0.1820    0.2789    0.2261    0.4568    0.1658    0.4862    0.1197    0.1529    0.1086    0.3525    0.1225    0.1057    0.2628    0.2197    0.4902    0.1760    0.3665    0.2411    0.1092    0.4911    0.1914    0.1182    0.3742
    0.1320    0.3658    0.1016    0.2159    0.3250    0.0930    0.3828    0.4593    0.4098    0.4640    0.3906    0.1789    0.4090    0.1879    0.2333    0.1869    0.4436    0.1210    0.4328    0.0966    0.1914    0.4469    0.1146    0.3296    0.2099    0.0977    0.1136    0.3288    0.2227    0.2751    0.1051    0.3514    0.0916    0.4277    0.3708    0.4496    0.1650    0.4224    0.4555    0.2375    0.2822    0.0899    0.3369    0.1557    0.2591    0.4483    0.0895    0.3164    0.4307    0.1508

exitflag = 
    SolverConvergedSuccessfully

output = struct with fields:
         iterations: 16
          funccount: 282
             volume: 1.8535
    averagedistance: 0.0089
             spread: 0.2737
      maxconstraint: 0
            message: 'Pareto set found that satisfies the constraints. ...'
           rngstate: [1x1 struct]
             solver: 'paretosearch'

The paretosearch solver converges in 16 iterations to a feasible solution. Plot the solution.

paretoplot(sol)

Figure contains an axes object. The axes object with title Pareto Front contains 4 objects of type text, scatter.

Choose an arbitrary point to examine in the plot using Data Tips:

arbitrarydatatip.png

arbitrary.png

The pictured point is at index 48. Examine solution 48.

arbitrarysol = sol(48)
arbitrarysol = 
  OptimizationValues with properties:

   Variables properties:
            x: 0.4375

   Objective properties:
    Objective: [2x1 double]

   Constraints properties:
         con1: -0.3086
         con2: -0.0914

The constraint values are negative, meaning the pictured point is feasible.

arbitrarysol.Objective
ans = 2×1

    0.1914
    0.3164

The objective values match the values in the Data Tips.

Limitations

  • OptimizationValues objects support horizontal concatenation only. In other words, you can have only row vectors of OptimizationValues objects.

Version History

Introduced in R2022a