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intfminbnd

Solves a multi-variable optimization problem on a bounded interval

Calling Sequence

xopt = intfminbnd(f,intcon,x1,x2)
xopt = intfminbnd(f,intcon,x1,x2,options)
[xopt,fopt] = intfminbnd(.....)
[xopt,fopt,exitflag]= intfminbnd(.....)
[xopt,fopt,exitflag,output]=intfminbnd(.....)
[xopt,fopt,exitflag,gradient,hessian]=intfminbnd(.....)

Parameters

f :

a function, representing the objective function of the problem

x1 :

a vector, containing the lower bound of the variables.

x2 :

a vector, containing the upper bound of the variables.

intcon :

a vector of integers, represents which variables are constrained to be integers

options :

a list, containing the option for user to specify. See below for details.

xopt :

a vector of doubles, containing the the computed solution of the optimization problem.

fopt :

a scalar of double, containing the the function value at x.

exitflag :

a scalar of integer, containing the flag which denotes the reason for termination of algorithm. See below for details.

gradient :

a vector of doubles, containing the Objective's gradient of the solution.

hessian :

a matrix of doubles, containing the Objective's hessian of the solution.

Description

Search the minimum of a multi-variable function on bounded interval specified by : Find the minimum of f(x) such that

The routine calls Bonmin for solving the Bounded Optimization problem, Bonmin is a library written in C++.

The options allows the user to set various parameters of the Optimization problem. It should be defined as type "list" and contains the following fields.

The exitflag allows to know the status of the optimization which is given back by Ipopt.

For more details on exitflag see the Bonmin documentation, go to http://www.coin-or.org/Bonmin

Examples

//Find x in R^6 such that it minimizes:
//f(x)= sin(x1) + sin(x2) + sin(x3) + sin(x4) + sin(x5) + sin(x6)
//-2 <= x1,x2,x3,x4,x5,x6 <= 2
//Objective function to be minimised
function y=f(x)
y=0
for i =1:6
y=y+sin(x(i));
end
endfunction
//Variable bounds
x1 = [-2, -2, -2, -2, -2, -2];
x2 = [2, 2, 2, 2, 2, 2];
intcon = [2 3 4]
//Options
options=list("MaxIter",[1500],"CpuTime", [100])
[x,fval] =intfminbnd(f ,intcon, x1, x2, options)
// Press ENTER to continue

Examples

//Find x in R such that it minimizes:
//f(x)= 1/x^2
//0 <= x <= 1000
//Objective function to be minimised
function y=f(x)
y=1/x^2;
endfunction
//Variable bounds
x1 = [0];
x2 = [1000];
intcon = [1];
[x,fval,exitflag,output,lambda] =intfminbnd(f,intcon , x1, x2)
// Press ENTER to continue

Examples

//The below problem is an unbounded problem:
//Find x in R^2 such that it minimizes:
//f(x)= -[(x1-1)^2 + (x2-1)^2]
//-inf <= x1,x2 <= inf
//Objective function to be minimised
function y=f(x)
y=-((x(1)-1)^2+(x(2)-1)^2);
endfunction
//Variable bounds
x1 = [-%inf , -%inf];
x2 = [ %inf , %inf];
//Options
options=list("MaxIter",[1500],"CpuTime", [100])
[x,fval,exitflag,output,lambda] =intfminbnd(f,intcon, x1, x2, options)

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