"The steady state contains NaN or Inf "


I have the following codes for the below system of equations which is hugely non-linear and I cannot do the linearization by hand. I did the codes in dynare in two versions. One using the exp(var) for the variables and the other one just by same equations.

However, i get always that “The steady state contains NaN or Inf” what would be the problem?
(The upload of the problem is below)


ex4.mod (1.9 KB)

ex5.mod (2.0 KB)


The upload of the problem:

the problem:
project (2).pdf (135.2 KB)
There is an integral in the model I could not write it and calculate in the dynare. so just for making it compatible and have a solution at least I assumed it as a parameter in my codes. I would be glad if someone also help me for that.


Please do not prematurely do an exp()-substitution. Use the model in levels and provide sensible initial values. You use 0 for most initval, but that is not allowed for e.g. capital.
Also, recheck all your equations. Steady state finding has a problem with b and k in particular. Check whether the equations containing those are correct.


Thanks for response.
I recheck my equations, and also simplified in terms of notations, ( I also attached the equations as well)

I have no clue for my initial values. I tried several initval but none of them changed the final answer to be different from NaN. Do you have any suggestions.
exx.mod (1.4 KB)

project (8).pdf (141.8 KB)


Initialize ALL variables. If you put


before steady, you will see that the starting values for various equations are still invalid.


Could you elaborate on that? what does the command “resid” do?


See the Dynare manual.


Hi everyone,
It seem`s like I have the same problem:
I can’t count initial values by hand and ask Dynare to do this, and get an error:
Error using print_info (line 90)
The steady state contains NaN or Inf
What should I do with this?
HA1.mod (1.6 KB)


Your initval block uses an implicit equation:


N appears both on the left and right of the equal sign. That cannot work. Either the equation is wrong or you need to call a solver.