evaluate
Evaluate optimization expression or objectives and constraints in problem
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, equation problem, or optimization constraint at a set of
points.
Tip
For the full workflow, see Problem-Based Optimization Workflow or Problem-Based Workflow for Solving Equations.
Examples
Evaluate Optimization Expression at Point
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
Evaluate Constraint Expressions
Create two optimization variables x
and y
and a 3-by-2 constraint expression in those variables.
x = optimvar("x"); y = optimvar("y"); cons = optimconstr(3,2); cons(1,1) = x^2 + y^2/4 <= 2; cons(1,2) = x^4 - y^4 <= -x^2 - y^2; cons(2,1) = x^2*3 + y^2 <= 2; cons(2,2) = x + y <= 3; cons(3,1) = x*y + x^2 + y^2 <= 5; cons(3,2) = x^3 + y^3 <= 8;
Evaluate the constraint expressions at the point , . The value of an expression L <= R
is L - R
.
x0.x = 1; x0.y = -1; val = evaluate(cons,x0)
val = 3×2
-0.7500 2.0000
2.0000 -3.0000
-4.0000 -8.0000
Evaluate Objective Function at Solution
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
Evaluate Optimization Problem Values
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.
Evaluate Expressions in Equation
Create a set of equations in two optimization variables.
x = optimvar("x"); y = optimvar("y"); prob = eqnproblem; prob.Equations.eq1 = x^2 + y^2/4 == 2; prob.Equations.eq2 = x^2/4 + 2*y^2 == 2;
Solve the system of equation starting from .
x0.x = 1; x0.y = 1/2; sol = solve(prob,x0)
Solving problem using fsolve. Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
sol = struct with fields:
x: 1.3440
y: 0.8799
Evaluate the equations at the points x0
and sol
.
vars = optimvalues(prob,x=[x0.x sol.x],y=[x0.y sol.y]); vals = evaluate(prob,vars)
vals = 1x2 OptimizationValues vector with properties: Variables properties: x: [1 1.3440] y: [0.5000 0.8799] Equation properties: eq1: [0.9375 8.4322e-10] eq2: [1.2500 6.7431e-09]
The first point, x0
, has nonzero values for both equations eq1
and eq2
. The second point, sol
, has nearly zero values of these equations, as expected for a solution.
Find the degree of equation satisfaction using issatisfied
.
[satisfied details] = issatisfied(prob,vars)
satisfied = 1x2 logical array
0 1
details = 1x2 OptimizationValues vector with properties: Variables properties: x: [1 1.3440] y: [0.5000 0.8799] Equation properties: eq1: [0 1] eq2: [0 1]
The first point, x0
, is not a solution, and satisfied
is 0
for that point. The second point, sol
, is a solution, and satisfied
is 1
for that point. The equation properties show that neither equation is satisfied at the first point, and both are satisfied at the second point.
Input Arguments
expr
— Optimization expression
OptimizationExpression
object
Optimization expression, specified as an OptimizationExpression
object.
Example: expr = 5*x+3
, where x
is an
OptimizationVariable
pt
— Values of variables in expression
structure
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 inpt
.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 inexpr
.
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
cons
— Constraint
OptimizationConstraint
object | OptimizationEquality
object | OptimizationInequality
object
Constraint, specified as an OptimizationConstraint
object, an OptimizationEquality
object, or an OptimizationInequality
object. evaluate
applies to these constraint objects only for a point specified as a structure, not as an OptimizationValues
object.
Example: cons = expr1 <= expr2
, where expr1
and expr2
are optimization expressions
prob
— Object for evaluation
OptimizationProblem
object | EquationProblem
object
Object for evaluation, specified as an OptimizationProblem
object or an EquationProblem
object. 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)
pts
— Points to evaluate for prob
structure | OptimizationValues
object
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 inprob
.The values in
pts
must be numeric arrays of the same size as the corresponding variables inprob
.
Note
Currently, pts
can be an OptimizationValues
object only when prob
is an EquationProblem
object or an OptimizationProblem
object.
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
val
— Evaluation result
double | OptimizationValues
object
Evaluation result, returned as a double or an OptimizationValues
object.
When the first input argument is an expression or constraint,
val
is returned as a double array of the same size as the expression or constraint, and contains its numeric values atpt
.When the first input argument is an
OptimizationProblem
object orEquationProblem
object,val
is anOptimizationValues
object.val
contains the values of the objective and constraints or equations inprob
evaluated at the points inpts
. Ifpts
containsN
points, thenval
has size 1-by-N
. For example, ifprob
contains a constraintcon
of size 2-by-3, andpts
is anOptimizationValues
object withN
= 5 points, thenval
has size 1-by-5, andval.Constraints.con
has size 2-by-3-by-5.
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
Constraint Expression Values
For a constraint expression at a point pt
:
If the constraint is
L <= R
, the constraint value isevaluate(L,pt)
–evaluate(R,pt)
.If the constraint is
L >= R
, the constraint value isevaluate(R,pt)
–evaluate(L,pt)
.If the constraint is
L == R
, the constraint value isabs(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 R2017bR2024b: Evaluate expressions in more objects
The evaluate
function now applies to the following objects:
For an example, see Evaluate Expressions in Equation.
R2024a: Evaluate objective and constraint expressions in optimization problems at multiple points
The evaluate
function now applies to objective and constraint
expressions in OptimizationProblem
objects. If the evaluation
points are passed as an OptimizationValues
object, then the
function evaluates the expressions at all points in the object. For an example, see
Evaluate Optimization Problem Values.
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