2015-07-22 15:08:25 +02:00
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function [X,exitflag]=disclyap_fast(G,V,tol,ch)
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2010-03-23 11:09:59 +01:00
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% function X=disclyap_fast(G,V,ch)
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2015-07-22 15:08:25 +02:00
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% Inputs:
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% - G [double] first input matrix
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% - V [double] second input matrix
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% - tol [scalar] tolerance criterion
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% - ch empty of non-empty if non-empty: check positive-definiteness
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% Outputs:
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% - X [double] solution matrix
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% - exitflag [scalar] 0 if solution is found, 1 otherwise
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%
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2010-03-23 11:09:59 +01:00
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% Solve the discrete Lyapunov Equation
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% X=G*X*G'+V
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% Using the Doubling Algorithm
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%
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% If ch is defined then the code will check if the resulting X
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% is positive definite and generate an error message if it is not
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%
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% Joe Pearlman and Alejandro Justiniano
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% 3/5/2005
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2010-10-11 17:16:52 +02:00
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2015-07-22 15:08:25 +02:00
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% Copyright (C) 2010-2015 Dynare Team
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2010-10-11 17:16:52 +02:00
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%
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% This file is part of Dynare.
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%
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% Dynare is free software: you can redistribute it and/or modify
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% it under the terms of the GNU General Public License as published by
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% the Free Software Foundation, either version 3 of the License, or
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% (at your option) any later version.
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%
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% Dynare is distributed in the hope that it will be useful,
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% but WITHOUT ANY WARRANTY; without even the implied warranty of
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% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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% GNU General Public License for more details.
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%
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% You should have received a copy of the GNU General Public License
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% along with Dynare. If not, see <http://www.gnu.org/licenses/>.
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2012-04-20 19:23:00 +02:00
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if nargin <= 3 || isempty( ch ) == 1
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2010-03-23 11:09:59 +01:00
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flag_ch = 0;
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else
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flag_ch = 1;
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end
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s=size(G,1);
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2015-07-22 15:08:25 +02:00
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exitflag=0;
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2010-03-23 11:09:59 +01:00
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2012-04-20 19:23:00 +02:00
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%tol = 1e-16;
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2010-03-23 11:09:59 +01:00
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P0=V;
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A0=G;
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matd=1;
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2015-07-22 15:08:25 +02:00
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iter=1;
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while matd > tol && iter< 2000
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2010-03-23 11:09:59 +01:00
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P1=P0+A0*P0*A0';
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A1=A0*A0;
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matd=max( max( abs( P1 - P0 ) ) );
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P0=P1;
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A0=A1;
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2015-07-22 15:08:25 +02:00
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iter=iter+1;
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2010-03-23 11:09:59 +01:00
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end
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2015-07-22 15:08:25 +02:00
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if iter==5000
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X=NaN(P0);
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exitflag=1;
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return
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end
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2010-03-23 11:09:59 +01:00
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clear A0 A1 P1;
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X=(P0+P0')/2;
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% Check that X is positive definite
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if flag_ch==1
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[C,p]=chol(X);
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if p ~= 0
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2015-07-22 15:08:25 +02:00
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exitflag=1;
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2010-03-23 11:09:59 +01:00
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error('X is not positive definite')
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end
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end
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