### 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.`
parent 33053a73
 ... @@ -42,6 +42,24 @@ nn = size(x,1); ... @@ -42,6 +42,24 @@ nn = size(x,1); % checking initial values % checking initial values if jacobian_flag if jacobian_flag [fvec,fjac] = feval(func,x,varargin{:}); [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 else fvec = feval(func,x,varargin{:}); fvec = feval(func,x,varargin{:}); fjac = zeros(nn,nn) ; fjac = zeros(nn,nn) ; ... ...
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