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Solve minimax constraint problem
fminimax
seeks a point that minimizes the maximum of a set of
objective functions.
The problem includes any type of constraint. In detail, fminimax
seeks the minimum of a problem specified by
$$\underset{x}{\mathrm{min}}\underset{i}{\mathrm{max}}{F}_{i}(x)\text{suchthat}\{\begin{array}{c}c(x)\le 0\\ ceq(x)=0\\ A\cdot x\le b\\ Aeq\cdot x=beq\\ lb\le x\le ub\end{array}$$
where b and beq are vectors, A and Aeq are matrices, and c(x), ceq(x), and F(x) are functions that return vectors. F(x), c(x), and ceq(x) can be nonlinear functions.
x, lb, and ub can be passed as vectors or matrices; see Matrix Arguments.
You can also solve maxmin problems with fminimax
, using the
identity
$$\underset{x}{\mathrm{max}}\underset{i}{\mathrm{min}}{F}_{i}(x)=\underset{x}{\mathrm{min}}\underset{i}{\mathrm{max}}\left({F}_{i}(x)\right).$$
You can solve problems of the form
$$\underset{x}{\mathrm{min}}\underset{i}{\mathrm{max}}\left{F}_{i}(x)\right$$
by using the AbsoluteMaxObjectiveCount
option; see Solve Minimax Problem Using Absolute Value of One Objective.
x = fminimax(fun,x0)
x = fminimax(fun,x0,A,b)
x = fminimax(fun,x0,A,b,Aeq,beq)
x = fminimax(fun,x0,A,b,Aeq,beq,lb,ub)
x = fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)
x = fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
x = fminimax(problem)
[x,fval]
= fminimax(___)
[x,fval,maxfval,exitflag,output]
= fminimax(___)
[x,fval,maxfval,exitflag,output,lambda]
= fminimax(___)
starts at x
= fminimax(fun
,x0
)x0
and finds a minimax solution x
to the
functions described in fun
.
Passing Extra Parameters explains how to pass extra parameters to the objective functions and nonlinear constraint functions, if necessary.
solves the minimax problem subject to the bounds
x
= fminimax(fun
,x0
,A
,b
,Aeq
,beq
,lb
,ub
)lb
≤ x
≤ ub
.
If no equalities exist, set Aeq = []
and beq = []
. If
x(i)
is unbounded below, set lb(i) = –Inf
; if
x(i)
is unbounded above, set ub(i) = Inf
.
If the specified input bounds for a problem are inconsistent, the output
x
is x0
and the output fval
is []
.
solves the minimax problem for x
= fminimax(problem
)problem
, where problem
is a structure described in problem
. Create
the problem
structure by exporting a problem from the Optimization app,
as described in Exporting Your Work.
sin
and cos
Create a plot of the sin
and cos
functions and their maximum over the interval [–pi,pi]
.
t = linspace(pi,pi); plot(t,sin(t),'r') hold on plot(t,cos(t),'b'); plot(t,max(sin(t),cos(t)),'ko') legend('sin(t)','cos(t)','max(sin(t),cos(t))','Location','NorthWest')
The plot shows two local minima of the maximum, one near 1, and the other near –2. Find the minimum near 1.
fun = @(x)[sin(x);cos(x)]; x0 = 1; x1 = fminimax(fun,x0)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x1 = 0.7854
Find the minimum near –2.
x0 = 2; x2 = fminimax(fun,x0)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x2 = 2.3562
The objective functions for this example are linear plus constants. For a description and plot of the objective functions, see Compare fminimax and fminunc.
Set the objective functions as three linear functions of the form $$dot(x,v)+{v}_{0}$$ for three vectors $$v$$ and three constants $${v}_{0}$$.
a = [1;1]; b = [1;1]; c = [0;1]; a0 = 2; b0 = 3; c0 = 4; fun = @(x)[x*a+a0,x*b+b0,x*c+c0];
Find the minimax point subject to the inequality x(1) + 3*x(2) <= –4
.
A = [1,3]; b = 4; x0 = [1,2]; x = fminimax(fun,x0,A,b)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
5.8000 0.6000
The objective functions for this example are linear plus constants. For a description and plot of the objective functions, see Compare fminimax and fminunc.
Set the objective functions as three linear functions of the form $$dot(x,v)+{v}_{0}$$ for three vectors $$v$$ and three constants $${v}_{0}$$.
a = [1;1]; b = [1;1]; c = [0;1]; a0 = 2; b0 = 3; c0 = 4; fun = @(x)[x*a+a0,x*b+b0,x*c+c0];
Set bounds that –2 <= x(1) <= 2
and –1 <= x(2) <= 1
and solve the minimax problem starting from [0,0]
.
lb = [2,1];
ub = [2,1];
x0 = [0,0];
A = []; % No linear constraints
b = [];
Aeq = [];
beq = [];
[x,fval] = fminimax(fun,x0,A,b,Aeq,beq,lb,ub)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
0.0000 1.0000
fval = 1×3
3.0000 2.0000 3.0000
In this case, the solution is not unique. Many points satisfy the constraints and have the same minimax value. Plot the surface representing the maximum of the three objective functions, and plot a red line showing the points that have the same minimax value.
[X,Y] = meshgrid(linspace(2,2),linspace(1,1)); Z = max(fun([X(:),Y(:)]),[],2); Z = reshape(Z,size(X)); surf(X,Y,Z,'LineStyle','none') view(118,28) hold on line([2,0],[1,1],[3,3],'Color','r','LineWidth',8) hold off
The objective functions for this example are linear plus constants. For a description and plot of the objective functions, see Compare fminimax and fminunc.
Set the objective functions as three linear functions of the form $$dot(x,v)+{v}_{0}$$ for three vectors $$v$$ and three constants $${v}_{0}$$.
a = [1;1]; b = [1;1]; c = [0;1]; a0 = 2; b0 = 3; c0 = 4; fun = @(x)[x*a+a0,x*b+b0,x*c+c0];
The unitdisk
function represents the nonlinear inequality constraint $$\Vert x{\Vert}^{2}\le 1$$.
type unitdisk
function [c,ceq] = unitdisk(x) c = x(1)^2 + x(2)^2  1; ceq = [];
Solve the minimax problem subject to the unitdisk
constraint, starting from x0 = [0,0]
.
x0 = [0,0];
A = []; % No other constraints
b = [];
Aeq = [];
beq = [];
lb = [];
ub = [];
nonlcon = @unitdisk;
x = fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
0.0000 1.0000
fminimax
can minimize the maximum of either $${F}_{i}(x)$$ or $${F}_{i}(x)$$ for the first several values of $$i$$ by using the AbsoluteMaxObjectiveCount
option. To minimize the absolute values of $$k$$ of the objectives, arrange the objective function values so that $${F}_{1}(x)$$ through $${F}_{k}(x)$$ are the objectives for absolute minimization, and set the AbsoluteMaxObjectiveCount
option to k
.
In this example, minimize the maximum of sin
and cos
, specify sin
as the first objective, and set AbsoluteMaxObjectiveCount
to 1.
fun = @(x)[sin(x),cos(x)]; options = optimoptions('fminimax','AbsoluteMaxObjectiveCount',1); x0 = 1; A = []; % No constraints b = []; Aeq = []; beq = []; lb = []; ub = []; nonlcon = []; x1 = fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x1 = 0.7854
Try starting from x0 = –2
.
x0 = 2; x2 = fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x2 = 3.1416
Plot the function.
t = linspace(pi,pi); plot(t,max(abs(sin(t)),cos(t)))
To see the effect of the AbsoluteMaxObjectiveCount
option, compare this plot to the plot in the example Minimize Maximum of sin and cos.
Obtain both the location of the minimax point and the value of the objective functions. For a description and plot of the objective functions, see Compare fminimax and fminunc.
Set the objective functions as three linear functions of the form $$dot(x,v)+{v}_{0}$$ for three vectors $$v$$ and three constants $${v}_{0}$$.
a = [1;1]; b = [1;1]; c = [0;1]; a0 = 2; b0 = 3; c0 = 4; fun = @(x)[x*a+a0,x*b+b0,x*c+c0];
Set the initial point to [0,0]
and find the minimax point and value.
x0 = [0,0]; [x,fval] = fminimax(fun,x0)
Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
2.5000 2.2500
fval = 1×3
1.7500 1.7500 1.7500
All three objective functions have the same value at the minimax point. Unconstrained problems typically have at least two objectives that are equal at the solution, because if a point is not a local minimum for any objective and only one objective has the maximum value, then the maximum objective can be lowered.
The objective functions for this example are linear plus constants. For a description and plot of the objective functions, see Compare fminimax and fminunc.
Set the objective functions as three linear functions of the form $$dot(x,v)+{v}_{0}$$ for three vectors $$v$$ and three constants $${v}_{0}$$.
a = [1;1]; b = [1;1]; c = [0;1]; a0 = 2; b0 = 3; c0 = 4; fun = @(x)[x*a+a0,x*b+b0,x*c+c0];
Find the minimax point subject to the inequality x(1) + 3*x(2) <= –4
.
A = [1,3]; b = 4; x0 = [1,2];
Set options for iterative display, and obtain all solver outputs.
options = optimoptions('fminimax','Display','iter'); Aeq = []; % No other constraints beq = []; lb = []; ub = []; nonlcon = []; [x,fval,maxfval,exitflag,output,lambda] =... fminimax(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
Objective Max Line search Directional Iter Fcount value constraint steplength derivative Procedure 0 4 0 6 1 9 5 0 1 0.981 2 14 4.889 0 1 0.302 Hessian modified twice 3 19 3.4 8.132e09 1 0.302 Hessian modified twice Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
x = 1×2
5.8000 0.6000
fval = 1×3
3.2000 3.4000 3.4000
maxfval = 3.4000
exitflag = 4
output = struct with fields:
iterations: 4
funcCount: 19
lssteplength: 1
stepsize: 6.0684e10
algorithm: 'activeset'
firstorderopt: []
constrviolation: 8.1323e09
message: '...'
lambda = struct with fields:
lower: [2x1 double]
upper: [2x1 double]
eqlin: [0x1 double]
eqnonlin: [0x1 double]
ineqlin: 0.2000
ineqnonlin: [0x1 double]
Examine the returned information:
Two objective function values are equal at the solution.
The solver converges in 4 iterations and 19 function evaluations.
The lambda.ineqlin
value is nonzero, indicating that the linear constraint is active at the solution.
fun
— Objective functionsObjective functions, specified as a function handle or function name.
fun
is a function that accepts a vector x
and
returns a vector F
, the objective functions evaluated at
x
. You can specify the function fun
as a
function handle for a function file:
x = fminimax(@myfun,x0,goal,weight)
where myfun
is a MATLAB^{®} function such as
function F = myfun(x) F = ... % Compute function values at x.
fun
can also be a function handle for an anonymous
function:
x = fminimax(@(x)sin(x.*x),x0,goal,weight);
If the userdefined values for x
and F
are
arrays, fminimax
converts them to vectors using linear indexing
(see Array Indexing (MATLAB)).
To minimize the worstcase absolute values of some elements of the vector
F(x) (that is, min{max
abs{F(x)} } ), partition those objectives into
the first elements of F and use optimoptions
to set the
AbsoluteMaxObjectiveCount
option to the number of these objectives. These
objectives must be partitioned into the first elements of the
vector F
returned by fun
. For an example, see
Solve Minimax Problem Using Absolute Value of One Objective.
Assume that the gradients of the objective functions can also be computed
and the SpecifyObjectiveGradient
option is
true
, as set by:
options = optimoptions('fminimax','SpecifyObjectiveGradient',true)
In this case, the function fun
must return, in the second output
argument, the gradient values G
(a matrix) at x
.
The gradient consists of the partial derivative dF/dx of each
F
at the point x
. If F
is a
vector of length m
and x
has length
n
, where n
is the length of
x0
, then the gradient G
of
F(x)
is an n
bym
matrix
where G(i,j)
is the partial derivative of F(j)
with respect to x(i)
(that is, the j
th column of
G
is the gradient of the j
th objective function
F(j)
). If you define F
as an array, then the
preceding discussion applies to F(:)
, the linear ordering of the
F
array. In any case, G
is a 2D matrix.
Setting SpecifyObjectiveGradient
to true
is
effective only when the problem has no nonlinear constraint, or when the problem has a
nonlinear constraint with SpecifyConstraintGradient
set to
true
. Internally, the objective is folded into the constraints,
so the solver needs both gradients (objective and constraint) supplied in order to
avoid estimating a gradient.
Data Types: char
 string
 function_handle
x0
— Initial pointInitial point, specified as a real vector or real array. Solvers use the number of elements in
x0
and the size of x0
to determine the number
and size of variables that fun
accepts.
Example: x0 = [1,2,3,4]
Data Types: double
A
— Linear inequality constraintsLinear inequality constraints, specified as a real matrix. A
is
an M
byN
matrix, where M
is
the number of inequalities, and N
is the number
of variables (number of elements in x0
). For
large problems, pass A
as a sparse matrix.
A
encodes the M
linear
inequalities
A*x <= b
,
where x
is the column vector of N
variables x(:)
,
and b
is a column vector with M
elements.
For example, to specify
x_{1} + 2x_{2} ≤
10
3x_{1} +
4x_{2} ≤ 20
5x_{1} +
6x_{2} ≤ 30,
enter these constraints:
A = [1,2;3,4;5,6]; b = [10;20;30];
Example: To specify that the x components sum to 1 or less, use A =
ones(1,N)
and b = 1
.
Data Types: double
b
— Linear inequality constraintsLinear inequality constraints, specified as a real vector. b
is
an M
element vector related to the A
matrix.
If you pass b
as a row vector, solvers internally
convert b
to the column vector b(:)
.
For large problems, pass b
as a sparse vector.
b
encodes the M
linear
inequalities
A*x <= b
,
where x
is the column vector of N
variables x(:)
,
and A
is a matrix of size M
byN
.
For example, to specify
x_{1} + 2x_{2} ≤
10
3x_{1} +
4x_{2} ≤ 20
5x_{1} +
6x_{2} ≤ 30,
enter these constraints:
A = [1,2;3,4;5,6]; b = [10;20;30];
Example: To specify that the x components sum to 1 or less, use A =
ones(1,N)
and b = 1
.
Data Types: double
Aeq
— Linear equality constraintsLinear equality constraints, specified as a real matrix. Aeq
is
an Me
byN
matrix, where Me
is
the number of equalities, and N
is the number of
variables (number of elements in x0
). For large
problems, pass Aeq
as a sparse matrix.
Aeq
encodes the Me
linear
equalities
Aeq*x = beq
,
where x
is the column vector of N
variables x(:)
,
and beq
is a column vector with Me
elements.
For example, to specify
x_{1} + 2x_{2} +
3x_{3} = 10
2x_{1} +
4x_{2} + x_{3} =
20,
enter these constraints:
Aeq = [1,2,3;2,4,1]; beq = [10;20];
Example: To specify that the x components sum to 1, use Aeq = ones(1,N)
and
beq = 1
.
Data Types: double
beq
— Linear equality constraintsLinear equality constraints, specified as a real vector. beq
is
an Me
element vector related to the Aeq
matrix.
If you pass beq
as a row vector, solvers internally
convert beq
to the column vector beq(:)
.
For large problems, pass beq
as a sparse vector.
beq
encodes the Me
linear
equalities
Aeq*x = beq
,
where x
is the column vector of N
variables
x(:)
, and Aeq
is a matrix of size
Me
byN
.
For example, to specify
x_{1} + 2x_{2} +
3x_{3} = 10
2x_{1} +
4x_{2} + x_{3} =
20,
enter these constraints:
Aeq = [1,2,3;2,4,1]; beq = [10;20];
Example: To specify that the x components sum to 1, use Aeq = ones(1,N)
and
beq = 1
.
Data Types: double
lb
— Lower boundsLower bounds, specified as a real vector or real array. If the number of elements in
x0
is equal to the number of elements in lb
,
then lb
specifies that
x(i) >= lb(i)
for all i
.
If numel(lb) < numel(x0)
, then lb
specifies
that
x(i) >= lb(i)
for 1 <=
i <= numel(lb)
.
If there are fewer elements in lb
than in x0
, solvers
issue a warning.
Example: To specify that all x components are positive, use lb =
zeros(size(x0))
.
Data Types: double
ub
— Upper boundsUpper bounds, specified as a real vector or real array. If the number of elements in
x0
is equal to the number of elements in ub
,
then ub
specifies that
x(i) <= ub(i)
for all i
.
If numel(ub) < numel(x0)
, then ub
specifies
that
x(i) <= ub(i)
for 1 <=
i <= numel(ub)
.
If there are fewer elements in ub
than in x0
, solvers
issue a warning.
Example: To specify that all x components are less than 1, use ub =
ones(size(x0))
.
Data Types: double
nonlcon
— Nonlinear constraintsNonlinear constraints, specified as a function handle or function name. nonlcon
is a function that accepts a vector or array x
and returns two arrays, c(x)
and ceq(x)
.
c(x)
is the array of nonlinear inequality constraints at x
. fminimax
attempts to satisfy
c(x) <= 0
for all entries of
c
.
ceq(x)
is the array of nonlinear equality constraints at x
. fminimax
attempts to satisfy
ceq(x) = 0
for all entries of
ceq
.
For example,
x = fminimax(@myfun,x0,...,@mycon)
where mycon
is a MATLAB function such as the
following:
function [c,ceq] = mycon(x) c = ... % Compute nonlinear inequalities at x. ceq = ... % Compute nonlinear equalities at x.
Suppose that the gradients of the constraints can also be computed and
the SpecifyConstraintGradient
option is true
, as
set by:
options = optimoptions('fminimax','SpecifyConstraintGradient',true)
In this case, the function nonlcon
must also return, in the third and
fourth output arguments, GC
, the gradient of c(x)
,
and GCeq
, the gradient of ceq(x)
. See Nonlinear Constraints for an explanation of how to “conditionalize” the
gradients for use in solvers that do not accept supplied gradients.
If nonlcon
returns a vector c
of m
components and x
has length n
, where
n
is the length of x0
, then the gradient
GC
of c(x)
is an
n
bym
matrix, where
GC(i,j)
is the partial derivative of c(j)
with
respect to x(i)
(that is, the j
th column of
GC
is the gradient of the j
th inequality
constraint c(j)
). Likewise, if ceq
has
p
components, the gradient GCeq
of
ceq(x)
is an n
byp
matrix,
where GCeq(i,j)
is the partial derivative of
ceq(j)
with respect to x(i)
(that is, the
j
th column of GCeq
is the gradient of the
j
th equality constraint ceq(j)
).
Setting SpecifyConstraintGradient
to true
is
effective only when SpecifyObjectiveGradient
is set to
true
. Internally, the objective is folded into the
constraint, so the solver needs both gradients (objective and constraint) supplied
in order to avoid estimating a gradient.
Because Optimization
Toolbox™ functions accept only inputs of type double
,
usersupplied objective and nonlinear constraint functions must return outputs of
type double
.
See Passing Extra Parameters for an explanation of how to parameterize the
nonlinear constraint function nonlcon
, if necessary.
Data Types: char
 function_handle
 string
options
— Optimization optionsoptimoptions
 structure such as optimset
returnsOptimization options, specified as the output of optimoptions
or a structure such as optimset
returns.
Some options are absent from the
optimoptions
display. These options appear in italics in the following
table. For details, see View Options.
For details about options that have different names for
optimset
, see Current and Legacy Option Name Tables.
Option  Description 

AbsoluteMaxObjectiveCount  Number of elements of F_{i}(x) for which to minimize the absolute value of F_{i}. See Solve Minimax Problem Using Absolute Value of One Objective. For 
ConstraintTolerance  Termination tolerance on the constraint violation (a positive
scalar). The default is For

Diagnostics  Display of diagnostic information about the function to be minimized
or solved. The choices are 
DiffMaxChange  Maximum change in variables for finitedifference gradients (a
positive scalar). The default is 
DiffMinChange  Minimum change in variables for finitedifference gradients (a
positive scalar). The default is 
 Level of display (see Iterative Display):

FiniteDifferenceStepSize 
Scalar or vector step size factor for finite differences. When
you set
sign′(x) = sign(x) except sign′(0) = 1 .
Central finite differences are
FiniteDifferenceStepSize expands to a vector. The default
is sqrt(eps) for forward finite differences, and eps^(1/3)
for central finite differences.
For 
FiniteDifferenceType  Type of finite differences used to estimate gradients, either
The algorithm is careful to obey bounds when estimating both types of finite differences. For example, it might take a backward difference, rather than a forward difference, to avoid evaluating at a point outside the bounds. For

FunctionTolerance  Termination tolerance on the function value (a positive scalar). The
default is For

FunValCheck  Check that signifies whether the objective function and constraint
values are valid. 
MaxFunctionEvaluations  Maximum number of function evaluations allowed (a positive integer).
The default is For

MaxIterations  Maximum number of iterations allowed (a positive integer). The
default is For

MaxSQPIter  Maximum number of SQP iterations allowed (a positive integer). The
default is 
MeritFunction  If this option is set to 
OptimalityTolerance 
Termination tolerance on the firstorder optimality (a positive
scalar). The default is For 
OutputFcn  One or more userdefined functions that an optimization function
calls at each iteration. Pass a function handle or a cell array of function
handles. The default is none ( 
PlotFcn  Plots showing various measures of progress while the algorithm
executes. Select from predefined plots or write your own. Pass a name,
function handle, or cell array of names or function handles. For custom plot
functions, pass function handles. The default is none
(
For information on writing a custom plot function, see Plot Function Syntax. For 
RelLineSrchBnd  Relative bound (a real nonnegative scalar value) on the line search
step length such that the total displacement in 
RelLineSrchBndDuration  Number of iterations for which the bound specified in

SpecifyConstraintGradient  Gradient for nonlinear constraint functions defined by the user. When
this option is set to For

SpecifyObjectiveGradient  Gradient for the objective function defined by the user. Refer to the
description of For 
StepTolerance  Termination tolerance on For

TolConSQP  Termination tolerance on the inner iteration SQP constraint violation
(a positive scalar). The default is 
TypicalX  Typical 
UseParallel  Option for using parallel computing. When this option is set to

Example: optimoptions('fminimax','PlotFcn','optimplotfval')
problem
— Problem structureProblem structure, specified as a structure with the fields in this table.
Field Name  Entry 

 Objective function fun 
 Initial point for x 
 Matrix for linear inequality constraints 
 Vector for linear inequality constraints 
 Matrix for linear equality constraints 
 Vector for linear equality constraints 
lb  Vector of lower bounds 
ub  Vector of upper bounds 
 Nonlinear constraint function 
 'fminimax' 
 Options created with optimoptions 
You must supply at least the objective
, x0
,
solver
, and options
fields in the
problem
structure.
The simplest way to obtain a problem
structure is to export the
problem from the Optimization app.
Data Types: struct
x
— SolutionSolution, returned as a real vector or real array. The size
of x
is the same as the size of x0
.
Typically, x
is a local solution to the problem
when exitflag
is positive. For information on
the quality of the solution, see When the Solver Succeeds.
fval
— Objective function values at solutionObjective function values at the solution, returned as a real array. Generally,
fval
= fun(x)
.
maxfval
— Maximum of objective function values at solutionMaximum of the objective function values at the solution, returned as a real scalar.
maxfval = max(fval(:))
.
exitflag
— Reason fminimax
stoppedReason fminimax
stopped, returned as an integer.
 Function converged to a solution

 Magnitude of the search direction was less than the specified
tolerance, and the constraint violation was less than

 Magnitude of the directional derivative was less than the
specified tolerance, and the constraint violation was less than

 Number of iterations exceeded

 Stopped by an output function or plot function 
 No feasible point was found. 
output
— Information about optimization processInformation about the optimization process, returned as a structure with the fields in this table.
iterations  Number of iterations taken 
funcCount  Number of function evaluations 
lssteplength  Size of the line search step relative to the search direction 
constrviolation  Maximum of the constraint functions 
stepsize  Length of the last displacement in

algorithm  Optimization algorithm used 
firstorderopt  Measure of firstorder optimality 
message  Exit message 
lambda
— Lagrange multipliers at solutionLagrange multipliers at the solution, returned as a structure with the fields in this table.
fminimax
solves a minimax problem by converting it into a goal
attainment problem, and then solving the converted goal attainment problem using
fgoalattain
. The conversion sets all goals to 0 and all weights to 1.
See Equation 1 in
Multiobjective Optimization Algorithms.
To run in parallel, set the 'UseParallel'
option to true
.
options = optimoptions('
solvername
','UseParallel',true)
For more information, see Using Parallel Computing in Optimization Toolbox.
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