Some element of Newton direction isn't finite

[Steph_Walras]

Hi, the running of my mod.file displays this error message:
Some element of Newton direction isn't finite. Jacobian maybe singular
or there is a problem with initial values

Error in ==> solve1 at 107
[x,f,fvec,check]=lnsrch1(xold,fold,g,p,stpmax,func,j1,j2,varargin{:});

Error in ==> dynare_solve at 150
[x,info]=solve1(func,x,j1(r(i):r(i+1)-1),j2(r(i):r(i+1)-1),jacobian_flag,
...

Error in ==> evaluate_steady_state at 66
[ys,check] = dynare_solve([M.fname '_static'],...

Error in ==> steady_ at 54
[steady_state,params,info] =
evaluate_steady_state(oo_.steady_state,M_,options_,oo_,~options_.steadystate.nocheck);
Error in ==> steady at 81
[steady_state,M_.params,info] = steady_(M_,options_,oo_);

Error in ==> regle at 183
steady;

Error in ==> dynare at 180
evalin('base',fname) ;

yet I compute these values relative to calibration. Thanks in advance.

[jpfeifer]

Your model is highly nonlinear and features a unit-root. That gives Dynare trouble. In the unstable version with

Code: Select all

steady(solve_algo=4,maxit=10000);


you will get the steady state, but you still have

Blanchard Kahn conditions are not satisfied: indeterminacy

[Steph_Walras]

Thank you for your held, I tried with steady(solve_algo=4,maxit=10000); but I didn't get the steady state. I still have the same error message. Please, what should I do now?

[jpfeifer]

This sounds like you did not use the Dynare unstable version, but Dynare 4.4.3. Please use the most recent Dynare snapshot.

[Steph_Walras]

Dear Jpfeiter,
I could get the steady state. But the problem is that it does not comply with restrictions on endogenous variables, which leads me to ask you to help me to elaborate FILE_steadystate.m. in which I can impose certain values to parameters compatible with the restrictions. Or is it possible to have an example of FILE_steadystate.m. for basic RBC. Thank you very much.

[jpfeifer]

What do you mean with restriction on endogenous variables? Your steady state is usually supposed to be unique. If it contradicts your intuition (e.g. labor is negative), then there must be a mistake in your model.

[Steph_Walras]

Thank you, I was able to solve the problem.