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|
// Copyright (C) 2015 - IIT Bombay - FOSSEE
//
// Author: Harpreet Singh
// Organization: FOSSEE, IIT Bombay
// Email: harpreet.mertia@gmail.com
// This file must be used under the terms of the CeCILL.
// This source file is licensed as described in the file COPYING, which
// you should have received as part of this distribution. The terms
// are also available at
// http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt
function [xopt,fopt,exitflag,output,lambda] = qpipoptmat (varargin)
// Solves a linear quadratic problem.
//
// Calling Sequence
// xopt = qpipoptmat(H,f)
// xopt = qpipoptmat(H,f,A,b)
// xopt = qpipoptmat(H,f,A,b,Aeq,beq)
// xopt = qpipoptmat(H,f,A,b,Aeq,beq,lb,ub)
// xopt = qpipoptmat(H,f,A,b,Aeq,beq,lb,ub,x0)
// xopt = qpipoptmat(H,f,A,b,Aeq,beq,lb,ub,x0,param)
// [xopt,fopt,exitflag,output,lamda] = qpipoptmat( ... )
//
// Parameters
// H : a symmetric matrix of doubles, represents coefficients of quadratic in the quadratic problem.
// f : a vector of doubles, represents coefficients of linear in the quadratic problem
// A : a vector of doubles, represents the linear coefficients in the inequality constraints
// b : a vector of doubles, represents the linear coefficients in the inequality constraints
// Aeq : a matrix of doubles, represents the linear coefficients in the equality constraints
// beq : a vector of doubles, represents the linear coefficients in the equality constraints
// LB : a vector of doubles, contains lower bounds of the variables.
// UB : a vector of doubles, contains upper bounds of the variables.
// x0 : a vector of doubles, contains initial guess of variables.
// param : a list containing the the parameters to be set.
// xopt : a vector of doubles, the computed solution of the optimization problem.
// fopt : a double, the function value at x.
// exitflag : Integer identifying the reason the algorithm terminated.
// output : Structure containing information about the optimization. Right now it contains number of iteration.
// lambda : Structure containing the Lagrange multipliers at the solution x (separated by constraint type).It contains lower, upper and linear equality, inequality constraints.
//
// Description
// Search the minimum of a constrained linear quadratic optimization problem specified by :
// find the minimum of f(x) such that
//
// <latex>
// \begin{eqnarray}
// &\mbox{min}_{x}
// & 1/2*x'*H*x + f'*x \\
// & \text{subject to} & A*x \leq b \\
// & & Aeq*x = beq \\
// & & lb \leq x \leq ub \\
// \end{eqnarray}
// </latex>
//
// We are calling IPOpt for solving the quadratic problem, IPOpt is a library written in C++.
//
// Examples
// //Find the value of x that minimize following function
// // f(x) = 0.5*x1^2 + x2^2 - x1*x2 - 2*x1 - 6*x2
// // Subject to:
// // x1 + x2 ≤ 2
// // –x1 + 2x2 ≤ 2
// // 2x1 + x2 ≤ 3
// // 0 ≤ x1, 0 ≤ x2.
// H = [1 -1; -1 2];
// f = [-2; -6];
// A = [1 1; -1 2; 2 1];
// b = [2; 2; 3];
// lb = [0; 0];
// ub = [%inf; %inf];
// [xopt,fopt,exitflag,output,lambda] = qpipoptmat(H,f,A,b,[],[],lb,ub)
// // Press ENTER to continue
//
// Examples
// //Find x in R^6 such that:
// Aeq= [1,-1,1,0,3,1;
// -1,0,-3,-4,5,6;
// 2,5,3,0,1,0];
// beq=[1; 2; 3];
// A= [0,1,0,1,2,-1;
// -1,0,2,1,1,0];
// b = [-1; 2.5];
// lb=[-1000; -10000; 0; -1000; -1000; -1000];
// ub=[10000; 100; 1.5; 100; 100; 1000];
// x0 = repmat(0,6,1);
// param = list("MaxIter", 300, "CpuTime", 100);
// //and minimize 0.5*x'*Q*x + p'*x with
// f=[1; 2; 3; 4; 5; 6]; H=eye(6,6);
// [xopt,fopt,exitflag,output,lambda]=qpipoptmat(H,f,A,b,Aeq,beq,lb,ub,[],param)
// Authors
// Keyur Joshi, Saikiran, Iswarya, Harpreet Singh
//To check the number of input and output argument
[lhs , rhs] = argn();
//To check the number of argument given by user
if ( rhs < 2 | rhs == 3 | rhs == 5 | rhs == 7 | rhs > 10 ) then
errmsg = msprintf(gettext("%s: Unexpected number of input arguments : %d provided while should be in the set of [2 4 6 8 9 10]"), "qpipoptmat", rhs);
error(errmsg)
end
H = varargin(1);
f = varargin(2);
nbVar = size(H,1);
if ( rhs<3 ) then
A = []
b = []
else
A = varargin(3);
b = varargin(4);
end
if ( rhs<5 ) then
Aeq = []
beq = []
else
Aeq = varargin(5);
beq = varargin(6);
end
if ( rhs<7 ) then
LB = repmat(-%inf,nbVar,1);
UB = repmat(%inf,nbVar,1);
else
LB = varargin(7);
UB = varargin(8);
end
if ( rhs<9 | size(varargin(9)) ==0 ) then
x0 = repmat(0,nbVar,1)
else
x0 = varargin(9);
end
if ( rhs<10 | size(varargin(10)) ==0 ) then
param = list();
else
param =varargin(10);
end
if (size(LB,2)==0) then
LB = repmat(-%inf,nbVar,1);
end
if (size(UB,2)==0) then
UB = repmat(%inf,nbVar,1);
end
if (size(f,2)==0) then
f = repmat(0,nbVar,1);
end
if (type(param) ~= 15) then
errmsg = msprintf(gettext("%s: param should be a list "), "qpipoptmat");
error(errmsg);
end
if (modulo(size(param),2)) then
errmsg = msprintf(gettext("%s: Size of parameters should be even"), "qpipoptmat");
error(errmsg);
end
options = list(..
"MaxIter" , [3000], ...
"CpuTime" , [600] ...
);
for i = 1:(size(param))/2
select param(2*i-1)
case "MaxIter" then
options(2*i) = param(2*i);
case "CpuTime" then
options(2*i) = param(2*i);
else
errmsg = msprintf(gettext("%s: Unrecognized parameter name ''%s''."), "qpipoptmat", param(2*i-1));
error(errmsg)
end
end
nbConInEq = size(A,1);
nbConEq = size(Aeq,1);
// Check if the user gives row vector
// and Changing it to a column matrix
if (size(f,2)== [nbVar]) then
f=f';
end
if (size(LB,2)== [nbVar]) then
LB = LB';
end
if (size(UB,2)== [nbVar]) then
UB = UB';
end
if (size(b,2)==nbConInEq) then
b = b';
end
if (size(beq,2)== nbConEq) then
beq = beq';
end
if (size(x0,2)== [nbVar]) then
x0=x0';
end
//Checking the H matrix which needs to be a symmetric matrix
if ( ~isequal(H,H')) then
errmsg = msprintf(gettext("%s: H is not a symmetric matrix"), "qpipoptmat");
error(errmsg);
end
//Check the size of f which should equal to the number of variable
if ( size(f,1) ~= [nbVar]) then
errmsg = msprintf(gettext("%s: The number of rows and columns in H must be equal the number of elements of f"), "qpipoptmat");
error(errmsg);
end
//Check the size of inequality constraint which should be equal to the number of variables
if ( size(A,2) ~= nbVar & size(A,2) ~= 0) then
errmsg = msprintf(gettext("%s: The number of columns in A must be the same as the number of elements of f"), "qpipoptmat");
error(errmsg);
end
//Check the size of equality constraint which should be equal to the number of variables
if ( size(Aeq,2) ~= nbVar & size(Aeq,2) ~= 0 ) then
errmsg = msprintf(gettext("%s: The number of columns in Aeq must be the same as the number of elements of f"), "qpipoptmat");
error(errmsg);
end
//Check the size of Lower Bound which should be equal to the number of variables
if ( size(LB,1) ~= nbVar) then
errmsg = msprintf(gettext("%s: The Lower Bound is not equal to the number of variables"), "qpipoptmat");
error(errmsg);
end
//Check the size of Upper Bound which should equal to the number of variables
if ( size(UB,1) ~= nbVar) then
errmsg = msprintf(gettext("%s: The Upper Bound is not equal to the number of variables"), "qpipoptmat");
error(errmsg);
end
//Check the size of constraints of Lower Bound which should equal to the number of constraints
if ( size(b,1) ~= nbConInEq & size(b,1) ~= 0) then
errmsg = msprintf(gettext("%s: The number of rows in A must be the same as the number of elementsof b"), "qpipoptmat");
error(errmsg);
end
//Check the size of constraints of Upper Bound which should equal to the number of constraints
if ( size(beq,1) ~= nbConEq & size(beq,1) ~= 0) then
errmsg = msprintf(gettext("%s: The number of rows in Aeq must be the same as the number of elements of beq"), "qpipoptmat");
error(errmsg);
end
//Check the size of initial of variables which should equal to the number of variables
if ( size(x0,1) ~= nbVar) then
warnmsg = msprintf(gettext("%s: Ignoring initial guess of variables as it is not equal to the number of variables"), "qpipoptmat");
warning(warnmsg);
end
//Check if the user gives a matrix instead of a vector
if ((size(f,1)~=1)& (size(f,2)~=1)) then
errmsg = msprintf(gettext("%s: f should be a vector"), "qpipoptmat");
error(errmsg);
end
if (size(LB,1)~=1)& (size(LB,2)~=1) then
errmsg = msprintf(gettext("%s: Lower Bound should be a vector"), "qpipoptmat");
error(errmsg);
end
if (size(UB,1)~=1)& (size(UB,2)~=1) then
errmsg = msprintf(gettext("%s: Upper Bound should be a vector"), "qpipoptmat");
error(errmsg);
end
if (nbConInEq) then
if ((size(b,1)~=1)& (size(b,2)~=1)) then
errmsg = msprintf(gettext("%s: Constraint Lower Bound should be a vector"), "qpipoptmat");
error(errmsg);
end
end
if (nbConEq) then
if (size(beq,1)~=1)& (size(beq,2)~=1) then
errmsg = msprintf(gettext("%s: Constraint should be a vector"), "qpipoptmat");
error(errmsg);
end
end
for i = 1:nbConInEq
if (b(i) == -%inf)
errmsg = msprintf(gettext("%s: Value of b can not be negative infinity"), "qpipoptmat");
error(errmsg);
end
end
for i = 1:nbConEq
if (beq(i) == -%inf)
errmsg = msprintf(gettext("%s: Value of beq can not be negative infinity"), "qpipoptmat");
error(errmsg);
end
end
//Converting it into ipopt format
f = f';
LB = LB';
UB = UB';
x0 = x0';
conMatrix = [Aeq;A];
nbCon = size(conMatrix,1);
conLB = [beq; repmat(-%inf,nbConInEq,1)]';
conUB = [beq;b]' ;
[xopt,fopt,status,iter,Zl,Zu,lmbda] = solveqp(nbVar,nbCon,H,f,conMatrix,conLB,conUB,LB,UB,x0,options);
xopt = xopt';
exitflag = status;
output = struct("Iterations" , []);
output.Iterations = iter;
lambda = struct("lower" , [], ..
"upper" , [], ..
"eqlin" , [], ..
"ineqlin" , []);
lambda.lower = Zl;
lambda.upper = Zu;
lambda.eqlin = lmbda(1:nbConEq);
lambda.ineqlin = lmbda(nbConEq+1:nbCon);
select status
case 0 then
printf("\nOptimal Solution Found.\n");
case 1 then
printf("\nMaximum Number of Iterations Exceeded. Output may not be optimal.\n");
case 2 then
printf("\nMaximum CPU Time exceeded. Output may not be optimal.\n");
case 3 then
printf("\nStop at Tiny Step\n");
case 4 then
printf("\nSolved To Acceptable Level\n");
case 5 then
printf("\nConverged to a point of local infeasibility.\n");
case 6 then
printf("\nStopping optimization at current point as requested by user.\n");
case 7 then
printf("\nFeasible point for square problem found.\n");
case 8 then
printf("\nIterates diverging; problem might be unbounded.\n");
case 9 then
printf("\nRestoration Failed!\n");
case 10 then
printf("\nError in step computation (regularization becomes too large?)!\n");
case 12 then
printf("\nProblem has too few degrees of freedom.\n");
case 13 then
printf("\nInvalid option thrown back by IPOpt\n");
case 14 then
printf("\nNot enough memory.\n");
case 15 then
printf("\nINTERNAL ERROR: Unknown SolverReturn value - Notify IPOPT Authors.\n");
else
printf("\nInvalid status returned. Notify the Toolbox authors\n");
break;
end
endfunction
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