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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,resnorm,residual,exitflag,output,lambda] = lsqlin (varargin)
// Solves a linear quadratic problem.
//
// Calling Sequence
// xopt = lsqlin(C,d,A,b)
// xopt = lsqlin(C,d,A,b,Aeq,beq)
// xopt = lsqlin(C,d,A,b,Aeq,beq,lb,ub)
// xopt = lsqlin(C,d,A,b,Aeq,beq,lb,ub,x0)
// xopt = lsqlin(C,d,A,b,Aeq,beq,lb,ub,x0,param)
// [xopt,resnorm,residual,exitflag,output,lambda] = lsqlin( ... )
//
// Parameters
// C : a matrix of double, represents the multiplier of the solution x in the expression C*x - d. Number of columns in C is equal to the number of elements in x.
// d : a vector of double, represents the additive constant term in the expression C*x - d. Number of elements in d is equal to the number of rows in C matrix.
// A : a vector of double, represents the linear coefficients in the inequality constraints
// b : a vector of double, represents the linear coefficients in the inequality constraints
// Aeq : a matrix of double, represents the linear coefficients in the equality constraints
// beq : a vector of double, represents the linear coefficients in the equality constraints
// lb : a vector of double, contains lower bounds of the variables.
// ub : a vector of double, contains upper bounds of the variables.
// x0 : a vector of double, contains initial guess of variables.
// param : a list containing the the parameters to be set.
// xopt : a vector of double, the computed solution of the optimization problem.
// resnorm : a double, objective value returned as the scalar value norm(C*x-d)^2.
// residual : a vector of double, solution residuals returned as the vector C*x-d.
// exitflag : Integer identifying the reason the algorithm terminated. It could be 0, 1 or 2 etc. i.e. Optimal, Maximum Number of Iterations Exceeded, CPU time exceeded. Other flags one can see in the lsqlin macro.
// output : Structure containing information about the optimization. This version only contains number of iterations.
// lambda : Structure containing the Lagrange multipliers at the solution x (separated by constraint type).It contains lower, upper bound multiplier and linear equality, inequality constraints.
//
// Description
// Search the minimum of a constrained linear least square problem specified by :
//
// <latex>
// \begin{eqnarray}
// &\mbox{min}_{x}
// & 1/2||C⋅x - d||_2^2 \\
// & \text{subject to} & A⋅x \leq b \\
// & & Aeq⋅x = beq \\
// & & lb \leq x \leq ub \\
// \end{eqnarray}
// </latex>
//
// The routine calls Ipopt for solving the linear least square problem, Ipopt is a library written in C++.
//
// Examples
// //A simple linear least square example
// C = [0.9501 0.7620 0.6153 0.4057
// 0.2311 0.4564 0.7919 0.9354
// 0.6068 0.0185 0.9218 0.9169
// 0.4859 0.8214 0.7382 0.4102
// 0.8912 0.4447 0.1762 0.8936];
// d = [0.0578
// 0.3528
// 0.8131
// 0.0098
// 0.1388];
// A = [0.2027 0.2721 0.7467 0.4659
// 0.1987 0.1988 0.4450 0.4186
// 0.6037 0.0152 0.9318 0.8462];
// b = [0.5251
// 0.2026
// 0.6721];
// [xopt,resnorm,residual,exitflag,output,lambda] = lsqlin(C,d,A,b)
// // Press ENTER to continue
//
// Examples
// //A basic example for equality, inequality constraints and variable bounds
// C = [0.9501 0.7620 0.6153 0.4057
// 0.2311 0.4564 0.7919 0.9354
// 0.6068 0.0185 0.9218 0.9169
// 0.4859 0.8214 0.7382 0.4102
// 0.8912 0.4447 0.1762 0.8936];
// d = [0.0578
// 0.3528
// 0.8131
// 0.0098
// 0.1388];
// A =[0.2027 0.2721 0.7467 0.4659
// 0.1987 0.1988 0.4450 0.4186
// 0.6037 0.0152 0.9318 0.8462];
// b =[0.5251
// 0.2026
// 0.6721];
// Aeq = [3 5 7 9];
// beq = 4;
// lb = -0.1*ones(4,1);
// ub = 2*ones(4,1);
// [xopt,resnorm,residual,exitflag,output,lambda] = lsqlin(C,d,A,b,Aeq,beq,lb,ub)
// Authors
// 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 < 4 | 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 [4 6 8 9 10]"), "lsqlin", rhs);
error(errmsg)
end
// Initializing all the values to empty matrix
C=[];
d=[];
A=[];
b=[];
Aeq=[];
beq=[];
lb=[];
ub=[];
x0=[];
C = varargin(1);
d = varargin(2);
A = varargin(3);
b = varargin(4);
nbVar = size(C,2);
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 (type(param) ~= 15) then
errmsg = msprintf(gettext("%s: param should be a list "), "lsqlin");
error(errmsg);
end
if (modulo(size(param),2)) then
errmsg = msprintf(gettext("%s: Size of parameters should be even"), "lsqlin");
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''."), "lsqlin", 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(d,2)== [nbVar]) then
d=d';
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
//Check the size of d which should equal to the number of variable
if ( size(d,1) ~= size(C,1)) then
errmsg = msprintf(gettext("%s: The number of rows in C must be equal the number of elements of d"), "lsqlin");
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 columns in C"), "lsqlin");
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 d"), "lsqlin");
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"), "lsqlin");
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"), "lsqlin");
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 elements of b"), "lsqlin");
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"), "lsqlin");
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"), "lsqlin");
warning(warnmsg);
x0 = repmat(0,nbVar,1);
end
//Check if the user gives a matrix instead of a vector
if ((size(d,1)~=1)& (size(d,2)~=1)) then
errmsg = msprintf(gettext("%s: d should be a vector"), "lsqlin");
error(errmsg);
end
if (size(lb,1)~=1)& (size(lb,2)~=1) then
errmsg = msprintf(gettext("%s: Lower Bound should be a vector"), "lsqlin");
error(errmsg);
end
if (size(ub,1)~=1)& (size(ub,2)~=1) then
errmsg = msprintf(gettext("%s: Upper Bound should be a vector"), "lsqlin");
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"), "lsqlin");
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"), "lsqlin");
error(errmsg);
end
end
for i = 1:nbConInEq
if (b(i) == -%inf)
errmsg = msprintf(gettext("%s: Value of b can not be negative infinity"), "lsqlin");
error(errmsg);
end
end
for i = 1:nbConEq
if (beq(i) == -%inf)
errmsg = msprintf(gettext("%s: Value of beq can not be negative infinity"), "lsqlin");
error(errmsg);
end
end
//Converting it into Quadratic Programming Problem
H = C'*C;
f = [-C'*d]';
op_add = d'*d;
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';
residual = -1*(C*xopt-d);
resnorm = residual'*residual;
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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