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Filter out cases where invalid starting value leads to NaN or Inf in Jacobian. 

Replaces cryptic message "Some element of Newton direction isn't finite. Jacobian maybe singular or there is a problem with initial values" with more expressive message.
time-shift
Johannes Pfeifer 2013-06-20 18:53:35 +02:00
parent 33053a7303
commit ab1cb25ed3
1 changed files with 18 additions and 0 deletions
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18
matlab/dynare_solve.m
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@ -42,6 +42,24 @@ nn = size(x,1);
% checking initial values
if jacobian_flag
[fvec,fjac] = feval(func,x,varargin{:});
if any(any(isinf(fjac) | isnan(fjac)))
[infrow,infcol]=find(isinf(fjac) | isnan(fjac));
M=evalin('base','M_'); %get variable names from workspace
fprintf('\nSTEADY: The Jacobian contains Inf or NaN. The problem arises from: \n\n')
for ii=1:length(infrow)
if infcol(ii)<=M.orig_endo_nbr
fprintf('STEADY: Derivative of Equation %d with respect to Variable %s (initial value of %s: %g) \n',infrow(ii),M.endo_names(infcol(ii),:),M.endo_names(infcol(ii),:),x(infcol(ii)))
else %auxiliary vars
orig_var_index=M.aux_vars(1,infcol(ii)-M.orig_endo_nbr).orig_index;
fprintf('STEADY: Derivative of Equation %d with respect to Variable %s (initial value of %s: %g) \n',infrow(ii),M.endo_names(orig_var_index,:),M.endo_names(orig_var_index,:),x(infcol(ii)))
end
end
fprintf('\nSTEADY: The problem most often occurs, because a variable with\n')
fprintf('STEADY: exponent smaller than 1 has been initialized to 0. Taking the derivative\n')
fprintf('STEADY: and evaluating it at the steady state then results in a division by 0.\n')
error('An element of the Jacobian is not finite or NaN')
end
else
fvec = feval(func,x,varargin{:});
fjac = zeros(nn,nn) ;