diff options
Diffstat (limited to 'newstructure/help/en_US/intfmincon.xml')
| -rw-r--r-- | newstructure/help/en_US/intfmincon.xml | 291 |
1 files changed, 0 insertions, 291 deletions
diff --git a/newstructure/help/en_US/intfmincon.xml b/newstructure/help/en_US/intfmincon.xml deleted file mode 100644 index a09a18a..0000000 --- a/newstructure/help/en_US/intfmincon.xml +++ /dev/null @@ -1,291 +0,0 @@ -<?xml version="1.0" encoding="UTF-8"?> - -<!-- - * - * This help file was generated from intfmincon.sci using help_from_sci(). - * - --> - -<refentry version="5.0-subset Scilab" xml:id="intfmincon" xml:lang="en" - xmlns="http://docbook.org/ns/docbook" - xmlns:xlink="http://www.w3.org/1999/xlink" - xmlns:svg="http://www.w3.org/2000/svg" - xmlns:ns3="http://www.w3.org/1999/xhtml" - xmlns:mml="http://www.w3.org/1998/Math/MathML" - xmlns:scilab="http://www.scilab.org" - xmlns:db="http://docbook.org/ns/docbook"> - - <refnamediv> - <refname>intfmincon</refname> - <refpurpose>Solves a constrainted multi-variable mixed integer non linear programming problem</refpurpose> - </refnamediv> - - -<refsynopsisdiv> - <title>Calling Sequence</title> - <synopsis> - xopt = intfmincon(f,x0,intcon,A,b) - xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq) - xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub) - xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub,nlc) - xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) - [xopt,fopt] = intfmincon(.....) - [xopt,fopt,exitflag]= intfmincon(.....) - [xopt,fopt,exitflag,gradient]=intfmincon(.....) - [xopt,fopt,exitflag,gradient,hessian]=intfmincon(.....) - - </synopsis> -</refsynopsisdiv> - -<refsection> - <title>Parameters</title> - <variablelist> - <varlistentry><term>f :</term> - <listitem><para> a function, representing the objective function of the problem</para></listitem></varlistentry> - <varlistentry><term>x0 :</term> - <listitem><para> a vector of doubles, containing the starting values of variables.</para></listitem></varlistentry> - <varlistentry><term>intcon :</term> - <listitem><para> a vector of integers, represents which variables are constrained to be integers</para></listitem></varlistentry> - <varlistentry><term>A :</term> - <listitem><para> a matrix of double, represents the linear coefficients in the inequality constraints A⋅x ≤ b.</para></listitem></varlistentry> - <varlistentry><term>b :</term> - <listitem><para> a vector of double, represents the linear coefficients in the inequality constraints A⋅x ≤ b.</para></listitem></varlistentry> - <varlistentry><term>Aeq :</term> - <listitem><para> a matrix of double, represents the linear coefficients in the equality constraints Aeq⋅x = beq.</para></listitem></varlistentry> - <varlistentry><term>beq :</term> - <listitem><para> a vector of double, represents the linear coefficients in the equality constraints Aeq⋅x = beq.</para></listitem></varlistentry> - <varlistentry><term>lb :</term> - <listitem><para> Lower bounds, specified as a vector or array of double. lb represents the lower bounds elementwise in lb ≤ x ≤ ub.</para></listitem></varlistentry> - <varlistentry><term>ub :</term> - <listitem><para> Upper bounds, specified as a vector or array of double. ub represents the upper bounds elementwise in lb ≤ x ≤ ub.</para></listitem></varlistentry> - <varlistentry><term>nlc :</term> - <listitem><para> a function, representing the Non-linear Constraints functions(both Equality and Inequality) of the problem. It is declared in such a way that non-linear inequality constraints are defined first as a single row vector (c), followed by non-linear equality constraints as another single row vector (ceq). Refer Example for definition of Constraint function.</para></listitem></varlistentry> - <varlistentry><term>options :</term> - <listitem><para> a list, containing the option for user to specify. See below for details.</para></listitem></varlistentry> - <varlistentry><term>xopt :</term> - <listitem><para> a vector of doubles, containing the the computed solution of the optimization problem.</para></listitem></varlistentry> - <varlistentry><term>fopt :</term> - <listitem><para> a scalar of double, containing the the function value at x.</para></listitem></varlistentry> - <varlistentry><term>exitflag :</term> - <listitem><para> a scalar of integer, containing the flag which denotes the reason for termination of algorithm. See below for details.</para></listitem></varlistentry> - <varlistentry><term>gradient :</term> - <listitem><para> a vector of doubles, containing the Objective's gradient of the solution.</para></listitem></varlistentry> - <varlistentry><term>hessian :</term> - <listitem><para> a matrix of doubles, containing the Objective's hessian of the solution.</para></listitem></varlistentry> - </variablelist> -</refsection> - -<refsection> - <title>Description</title> - <para> -Search the minimum of a mixed integer constrained optimization problem specified by : -Find the minimum of f(x) such that - </para> - <para> -<latex> -\begin{eqnarray} -&\mbox{min}_{x} -& f(x) \\ -& \text{subject to} & A*x \leq b \\ -& & Aeq*x \ = beq\\ -& & c(x) \leq 0\\ -& & ceq(x) \ = 0\\ -& & lb \leq x \leq ub \\ -& & x_i \in \!\, \mathbb{Z}, i \in \!\, I -\end{eqnarray} -</latex> - </para> - <para> -The routine calls Bonmin for solving the Bounded Optimization problem, Bonmin is a library written in C++. - </para> - <para> -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. -<itemizedlist> -<listitem>Syntax : options= list("IntegerTolerance", [---], "MaxNodes",[---], "MaxIter", [---], "AllowableGap",[---] "CpuTime", [---],"gradobj", "off", "hessian", "off" );</listitem> -<listitem>IntegerTolerance : a Scalar, a number with that value of an integer is considered integer..</listitem> -<listitem>MaxNodes : a Scalar, containing the Maximum Number of Nodes that the solver should search.</listitem> -<listitem>CpuTime : a Scalar, containing the Maximum amount of CPU Time that the solver should take.</listitem> -<listitem>AllowableGap : a Scalar, to stop the tree search when the gap between the objective value of the best known solution is reached.</listitem> -<listitem>MaxIter : a Scalar, containing the Maximum Number of Iteration that the solver should take.</listitem> -<listitem>gradobj : a string, to turn on or off the user supplied objective gradient.</listitem> -<listitem>hessian : a Scalar, to turn on or off the user supplied objective hessian.</listitem> -<listitem>Default Values : options = list('integertolerance',1d-06,'maxnodes',2147483647,'cputime',1d10,'allowablegap',0,'maxiter',2147483647,'gradobj',"off",'hessian',"off")</listitem> -</itemizedlist> - </para> - <para> -The exitflag allows to know the status of the optimization which is given back by Ipopt. -<itemizedlist> -<listitem>exitflag=0 : Optimal Solution Found </listitem> -<listitem>exitflag=1 : InFeasible Solution.</listitem> -<listitem>exitflag=2 : Objective Function is Continuous Unbounded.</listitem> -<listitem>exitflag=3 : Limit Exceeded.</listitem> -<listitem>exitflag=4 : User Interrupt.</listitem> -<listitem>exitflag=5 : MINLP Error.</listitem> -</itemizedlist> - </para> - <para> -For more details on exitflag see the Bonmin documentation, go to http://www.coin-or.org/Bonmin - </para> - <para> -</para> -</refsection> - -<refsection> - <title>Examples</title> - <programlisting role="example"><![CDATA[ -//Find x in R^2 such that it minimizes: -//f(x)= -x1 -x2/3 -//x0=[0,0] -//constraint-1 (c1): x1 + x2 <= 2 -//constraint-2 (c2): x1 + x2/4 <= 1 -//constraint-3 (c3): x1 - x2 <= 2 -//constraint-4 (c4): -x1/4 - x2 <= 1 -//constraint-5 (c5): -x1 - x2 <= -1 -//constraint-6 (c6): -x1 + x2 <= 2 -//constraint-7 (c7): x1 + x2 = 2 -//Objective function to be minimised -function [y,dy]=f(x) -y=-x(1)-x(2)/3; -dy= [-1,-1/3]; -endfunction -//Starting point, linear constraints and variable bounds -x0=[0 , 0]; -intcon = [1] -A=[1,1 ; 1,1/4 ; 1,-1 ; -1/4,-1 ; -1,-1 ; -1,1]; -b=[2;1;2;1;-1;2]; -Aeq=[1,1]; -beq=[2]; -lb=[]; -ub=[]; -nlc=[]; -//Options -options=list("GradObj", "on"); -//Calling Ipopt -[x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) -// Press ENTER to continue - - ]]></programlisting> -</refsection> - -<refsection> - <title>Examples</title> - <programlisting role="example"><![CDATA[ -//Find x in R^3 such that it minimizes: -//f(x)= x1*x2 + x2*x3 -//x0=[0.1 , 0.1 , 0.1] -//constraint-1 (c1): x1^2 - x2^2 + x3^2 <= 2 -//constraint-2 (c2): x1^2 + x2^2 + x3^2 <= 10 -//Objective function to be minimised -function [y,dy]=f(x) -y=x(1)*x(2)+x(2)*x(3); -dy= [x(2),x(1)+x(3),x(2)]; -endfunction -//Starting point, linear constraints and variable bounds -x0=[0.1 , 0.1 , 0.1]; -intcon = [2] -A=[]; -b=[]; -Aeq=[]; -beq=[]; -lb=[]; -ub=[]; -//Nonlinear constraints -function [c,ceq,cg,cgeq]=nlc(x) -c = [x(1)^2 - x(2)^2 + x(3)^2 - 2 , x(1)^2 + x(2)^2 + x(3)^2 - 10]; -ceq = []; -cg=[2*x(1) , -2*x(2) , 2*x(3) ; 2*x(1) , 2*x(2) , 2*x(3)]; -cgeq=[]; -endfunction -//Options -options=list("MaxIter", [1500], "CpuTime", [500], "GradObj", "on","GradCon", "on"); -//Calling Ipopt -[x,fval,exitflag,output] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) -// Press ENTER to continue - - ]]></programlisting> -</refsection> - -<refsection> - <title>Examples</title> - <programlisting role="example"><![CDATA[ -//The below problem is an unbounded problem: -//Find x in R^3 such that it minimizes: -//f(x)= -(x1^2 + x2^2 + x3^2) -//x0=[0.1 , 0.1 , 0.1] -// x1 <= 0 -// x2 <= 0 -// x3 <= 0 -//Objective function to be minimised -function y=f(x) -y=-(x(1)^2+x(2)^2+x(3)^2); -endfunction -//Starting point, linear constraints and variable bounds -x0=[0.1 , 0.1 , 0.1]; -intcon = [3] -A=[]; -b=[]; -Aeq=[]; -beq=[]; -lb=[]; -ub=[0,0,0]; -//Options -options=list("MaxIter", [1500], "CpuTime", [500]); -//Calling Ipopt -[x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,[],options) -// Press ENTER to continue - - ]]></programlisting> -</refsection> - -<refsection> - <title>Examples</title> - <programlisting role="example"><![CDATA[ -//The below problem is an infeasible problem: -//Find x in R^3 such that in minimizes: -//f(x)=x1*x2 + x2*x3 -//x0=[1,1,1] -//constraint-1 (c1): x1^2 <= 1 -//constraint-2 (c2): x1^2 + x2^2 <= 1 -//constraint-3 (c3): x3^2 <= 1 -//constraint-4 (c4): x1^3 = 0.5 -//constraint-5 (c5): x2^2 + x3^2 = 0.75 -// 0 <= x1 <=0.6 -// 0.2 <= x2 <= inf -// -inf <= x3 <= 1 -//Objective function to be minimised -function [y,dy]=f(x) -y=x(1)*x(2)+x(2)*x(3); -dy= [x(2),x(1)+x(3),x(2)]; -endfunction -//Starting point, linear constraints and variable bounds -x0=[1,1,1]; -intcon = [2] -A=[]; -b=[]; -Aeq=[]; -beq=[]; -lb=[0 0.2,-%inf]; -ub=[0.6 %inf,1]; -//Nonlinear constraints -function [c,ceq,cg,cgeq]=nlc(x) -c=[x(1)^2-1,x(1)^2+x(2)^2-1,x(3)^2-1]; -ceq=[x(1)^3-0.5,x(2)^2+x(3)^2-0.75]; -cg = [2*x(1),0,0;2*x(1),2*x(2),0;0,0,2*x(3)]; -cgeq = [3*x(1)^2,0,0;0,2*x(2),2*x(3)]; -endfunction -//Options -options=list("MaxIter", [1500], "CpuTime", [500], "GradObj", "on","GradCon", "on"); -//Calling Ipopt -[x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) -// Press ENTER to continue - ]]></programlisting> -</refsection> - -<refsection> - <title>Authors</title> - <simplelist type="vert"> - <member>Harpreet Singh</member> - </simplelist> -</refsection> -</refentry> |