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OptimizationInequality

Inequality constraints

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Description

An OptimizationInequality object contains an inequality constraint in terms of OptimizationVariable objects or OptimizationExpression objects. An inequality constraint uses the comparison operator <= or >=.

A single statement can represent an array of inequalities. For example, you can express the inequalities that each row of a matrix variable x sums to no more than one in this single statement:

constrsum = sum(x,2) <= 1

Use OptimizationInequality objects as constraints in an OptimizationProblem.

Tip

For the full workflow, see Problem-Based Optimization Workflow.

Creation

Create an inequality using optimization expressions with the comparison operator <= or >=.

Include inequalities in the Constraints property of an optimization problem by using dot notation.

prob = optimproblem;
x = optimvar('x',4,6);
SumLessThanOne = sum(x,2) <= 1;
prob.Constraints.SumLessThanOne = SumLessThanOne;

You can also create an empty optimization inequality by using optimineq or optimconstr. Typically, you then set the inequalities in a loop. For an example, see Create Inequalities in Loop. However, for the most efficient problem formulation, avoid setting inequalities in loops. See Create Efficient Optimization Problems.

Properties

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Index names, specified as a cell array of strings or character vectors. For information on using index names, see Named Index for Optimization Variables.

Data Types: cell

This property is read-only.

Optimization variables in the object, specified as a structure of OptimizationVariable objects.

Data Types: struct

Object Functions

evaluateEvaluate optimization expression or objectives and constraints in problem
infeasibilityConstraint violation at a point
issatisfiedConstraint satisfaction of an optimization problem at a set of points
showDisplay information about optimization object
writeSave optimization object description

Examples

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Open Live Script

Create a 4-by-6 optimization variable matrix named x. 

x = optimvar('x',4,6);

Create the inequalities that each row of x sums to no more than one.

constrsum = sum(x,2) <= 1
constrsum = 
  4x1 Linear OptimizationInequality array with properties:

    IndexNames: {{}  {}}
     Variables: [1x1 struct] containing 1 OptimizationVariable

  See inequality formulation with show.

View the inequalities.

show(constrsum)
(1, 1)

  x(1, 1) + x(1, 2) + x(1, 3) + x(1, 4) + x(1, 5) + x(1, 6) <= 1

(2, 1)

  x(2, 1) + x(2, 2) + x(2, 3) + x(2, 4) + x(2, 5) + x(2, 6) <= 1

(3, 1)

  x(3, 1) + x(3, 2) + x(3, 3) + x(3, 4) + x(3, 5) + x(3, 6) <= 1

(4, 1)

  x(4, 1) + x(4, 2) + x(4, 3) + x(4, 4) + x(4, 5) + x(4, 6) <= 1

To include the inequalities in an optimization problem, set a Constraints property to constrsum by using dot notation.

prob = optimproblem;
prob.Constraints.constrsum = constrsum
prob = 
  OptimizationProblem with properties:

       Description: ''
    ObjectiveSense: 'minimize'
         Variables: [1x1 struct] containing 1 OptimizationVariable
         Objective: [0x0 OptimizationExpression]
       Constraints: [1x1 struct] containing 1 OptimizationConstraint

  See problem formulation with show.
Open Live Script

Create the constraint that a two-element variable x must lie in the intersections of a number of disks whose centers and radii are in the arrays centers and radii.

x = optimvar('x',1,2);
centers = [1 -2;3 -4;-2 3];
radii = [6 7 8];
constr = optimineq(length(radii));
for i = 1:length(constr)
    constr(i) = sum((x - centers(i,:)).^2) <= radii(i)^2;
end

View the inequality expressions.

show(constr)
  arg_LHS <= arg_RHS

  where:

        arg1 = zeros(3, 1);
        arg1(1) = sum((x - extraParams{1}).^2);
        arg1(2) = sum((x - extraParams{2}).^2);
        arg1(3) = sum((x - extraParams{3}).^2);
        arg_LHS = arg1(:);
        arg1 = zeros(3, 1);
        arg1(1) = 36;
        arg1(2) = 49;
        arg1(3) = 64;
        arg_RHS = arg1(:);

    extraParams{1}:

     1    -2


  extraParams{2}:

     3    -4


  extraParams{3}:

    -2     3

Instead of using a loop, you can create the same constraints by using matrix operations on the variables.

constr2 = sum(([x;x;x] - centers).^2,2) <= radii'.^2;

Creating inequalities in a loop can be more time consuming than creating inequalities by using matrix operations.

Open Live Script

Check the value of an inequality constraint and satisfaction of an equality constraint at a point specified by a structure.

x = optimvar("x");
y = optimvar("y");
ineq = x^2 + y^2/4 <= 2;
eq = 4*x^2 + y^2 == 5;
x0.x = 1;
x0.y = 1;
evaluate(ineq,x0) % Value = L - R, where L =  x^2 + y^2/4 and R = 2
ans = 
-0.7500

issatisfied(eq,x0) % Checks whether 4*x^2 + y^2 = 5
ans = logical
   1

Version History

Introduced in R2019b

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See Also

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