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evaluate

Evaluate optimization expression or objectives and constraints in problem

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Syntax

val = evaluate(expr,pt)
val = evaluate(prob,pts)

Description

Use evaluate to find the numeric value of an optimization expression at a point, or to find the values of objective and constraint expressions in an optimization problem at a set of points.

Tip

For the full workflow, see Problem-Based Optimization Workflow or Problem-Based Workflow for Solving Equations.

example

val = evaluate(expr,pt) returns the value of the optimization expression expr at the value pt.

example

val = evaluate(prob,pts) returns the values of the objective and constraint functions in prob at the points in pts.

Examples

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Open Live Script

Create an optimization expression of two variables. 

x = optimvar("x",3,2);
y = optimvar("y",1,2);
expr = sum(x,1) - 2*y;

Evaluate the expression at a point.

xmat = [3,-1;
    0,1;
    2,6];
sol.x = xmat;
sol.y = [4,-3];
val = evaluate(expr,sol)
val = 1×2

    -3    12
Open Live Script

Solve a linear programming problem. 

x = optimvar('x');
y = optimvar('y');
prob = optimproblem;
prob.Objective = -x -y/3;
prob.Constraints.cons1 = x + y <= 2;
prob.Constraints.cons2 = x + y/4 <= 1;
prob.Constraints.cons3 = x - y <= 2;
prob.Constraints.cons4 = x/4 + y >= -1;
prob.Constraints.cons5 = x + y >= 1;
prob.Constraints.cons6 = -x + y <= 2;

sol = solve(prob)
Solving problem using linprog.

Optimal solution found.

sol = struct with fields:
    x: 0.6667
    y: 1.3333

Find the value of the objective function at the solution.

val = evaluate(prob.Objective,sol)
val = -1.1111
Open Live Script

Create an optimization problem with several linear and nonlinear constraints.

x = optimvar("x");
y = optimvar("y");
obj = (10*(y - x^2))^2 + (1 - x)^2;
cons1 = x^2 + y^2 <= 1;
cons2 = x + y >= 0;
cons3 = y <= sin(x);
cons4 = 2*x + 3*y <= 2.5;
prob = optimproblem(Objective=obj);
prob.Constraints.cons1 = cons1;
prob.Constraints.cons2 = cons2;
prob.Constraints.cons3 = cons3;
prob.Constraints.cons4 = cons4;

Create 100 test points randomly.

rng default % For reproducibility
xvals = randn(1,100);
yvals = randn(1,100);

Convert the points to an OptimizationValues object for the problem.

pts = optimvalues(prob,x=xvals,y=yvals);

Evaluate the objective and constraint functions at the points pts.

val = evaluate(prob,pts);

The objective function values are stored in val.Objective, and the constraint function values are stored in val.cons1 through val.cons4. Plot the log of 1 plus the objective function values.

figure
plot3(xvals,yvals,log(1 + val.Objective),"bo")

Plot the values of the constraints cons1 and cons4. Recall that constraints are satisfied when they evaluate to a nonpositive number. Plot the nonpositive values with circles and the positive values with x marks.

neg1 = val.cons1 <= 0;
pos1 = val.cons1 > 0;
neg4 = val.cons4 <= 0;
pos4 = val.cons4 > 0;
figure
plot3(xvals(neg1),yvals(neg1),val.cons1(neg1),"bo")
hold on
plot3(xvals(pos1),yvals(pos1),val.cons1(pos1),"rx")
plot3(xvals(neg4),yvals(neg4),val.cons4(neg4),"ko")
plot3(xvals(pos4),yvals(pos4),val.cons4(pos4),"gx")
hold off

As the last figure shows, evaluate enables you to calculate both the value and the feasibility of points. In contrast, issatisfied calculates only the feasibility.

Input Arguments

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Optimization expression, specified as an OptimizationExpression object.

Example: expr = 5*x+3, where x is an OptimizationVariable

Values of the variables in an expression, specified as a structure. The structure pt has the following requirements:

  • All variables in expr must match field names in pt.

  • The values of the matching field names must be numeric.

  • The sizes of the fields in pt must match the sizes of the corresponding variables in expr.

For example, pt can be the solution to an optimization problem, as returned by solve.

Example: pt.x = 3, pt.y = -5

Data Types: struct

Optimization problem, specified as an OptimizationProblem object. Create prob using optimproblem. The evaluate function evaluates the objectives and constraints in the properties of prob at the points in pts.

Example: prob = optimproblem(Objective=obj,Constraints=constr)

Points to evaluate for prob, specified as a structure or an OptimizationValues object.

  • The field names in pts must match the corresponding variable names in the objective and constraint expressions in prob.

  • The values in pts must be numeric arrays of the same size as the corresponding variables in prob.

If you use a structure for pts, then pts can contain only one point. In other words, if you want to evaluate multiple points simultaneously, pts must be an OptimizationValues object.

Example: pts = optimvalues(prob,x=xval,y=yval)

Output Arguments

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Evaluation result, returned as a double or as an OptimizationValues object. If val is a double, it contains the numeric value of the expression at pt. If val is an OptimizationValues object, it contains values for the objective and constraint functions in prob evaluated at the points in pts.

Warning

The problem-based approach does not support complex values in the following: an objective function, nonlinear equalities, and nonlinear inequalities. If a function calculation has a complex value, even as an intermediate value, the final result might be incorrect.

More About

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Constraint Expression Values

For a constraint expression at a point pt:

  • If the constraint is L <= R, the constraint value is evaluate(L,pt) – evaluate(R,pt).

  • If the constraint is L >= R, the constraint value is evaluate(R,pt) – evaluate(L,pt).

  • If the constraint is L == R, the constraint value is abs(evaluate(L,pt) – evaluate(R,pt)).

Generally, a constraint is considered to be satisfied (or feasible) at a point if the constraint value is less than or equal to a tolerance.

Version History

Introduced in R2017b

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See Also

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