TO continue debugging, try rbc_adam.mod (1.4 KB)
I fixed the equation for Z and replaced steady_state_model, which requires the full analytical steady state but initval. Now the Blanchard/Kahn conditions fail. Usually it’s a timing error.

Thanks - I tried to declare/define the variables before using them in the initval-block but the problems seems not to be solved?

initval;
Z = 1;
Yc = 1-calpha_c;
Rc = calpha_c*(1-calpha_c)^(1-calpha_c);
Qc = (cbeta*Rc)/((1-cbeta)*(1-cdelta_c));
Hi = (((1-calpha_i)*Qc)/(1-calpha_c)^(1-calpha_c))^(1/calpha_i);
Yi = Hi^(1-calpha_i);
Y = Yc + Qc*Yi;
C = 1-calpha_c;
Hc = 1-calpha_c;
Ri = ((calpha_i)/(1-calpha_i))*Hi*((1-calpha_c)^(1-calpha_c));
Inv = Qc*(((1-calpha_i)*Qc)/W)^((1-calpha_i)/calpha_i);
W = 1-calpha_c;
end;

Error using print_info (line 32) The Jacobian matrix evaluated at the steady state contains elements that are not real or are infinite.

Error in check (line 48)

print_info(info, 0, options);*

Error in rbc_adam_new.driver (line 255) oo_.dr.eigval = check(M_,options_,oo_);

Error in dynare (line 293) evalin(‘base’,[fname ‘.driver’]) ;

Thanks again for looking at it.
This type of error are usually a timing error, right?

Error using print_info (line 32) Blanchard & Kahn conditions are not satisfied: no stable equilibrium.

initval;
Z = 1;
Hc = 1-calpha_c;
Yc = (1-calpha_c)^(1-calpha_c);
W = (1-calpha_c)^(1-calpha_c);
C = (1-calpha_c)^(1-calpha_c);
Rc = calpha_c*(1-calpha_c)^(1-calpha_c);
Qc = (cbetaRc)/((1-cbeta)(1-cdelta_c));
Hi = (((1-calpha_i)Qc)/(1-calpha_c)^(1-calpha_c))^(1/calpha_i);
Yi = Hi^(1-calpha_i);
Inv = Qc(((1-calpha_i)Qc)/W)^((1-calpha_i)/calpha_i);
Y = Yc + QcYi;
Ri = ((calpha_i)/(1-calpha_i))Hi((1-calpha_c)^(1-calpha_c));
end;

I tried to change the timing for each of the variables relative to when they are declared/included, but i still cannot achieve a solution. Would it be possible for your to provide som insight/tips on where the timing of the variables goes wrong?

Thanks for your feedback. Do you think the model would be able to estimate the simulation by rearranging the timing of variables as it has declared values of steady states? Or should I look at the steady states equations again?

I have assumed it is equal to one as I found k_i equal to one in steady state and then equalised k_i = k_c as a rough assumption in order to find steady state values for the other variables

Yes. I have tried to declare the variables Ki and Kc in the model block hence i again receive issues with the steady state values

Residuals of the static equations:

Equation number 1 : 0 : C
Equation number 2 : 44.47 : Qc
Equation number 3 : 0 : Rc
Equation number 4 : 0 : Hc
Equation number 5 : 0 : W
Equation number 6 : 0 : Ri
Equation number 7 : 0 : Yc
Equation number 8 : 0 : Yi
Equation number 9 : 0 : Y
Equation number 10 : 0 : Inv
Equation number 11 : 0 : Hi
Equation number 12 : -3.2518 : Kc
Equation number 13 : 0 : Ki
Equation number 14 : 0 : X
Equation number 15 : 0 : 15

Error using print_info (line 32)
The steadystate file did not compute the steady state

Error in steady (line 102)
print_info(info,options_.noprint, options_);

Error in rbc_adam_new.driver (line 241)
steady;

Error in dynare (line 293)
evalin(‘base’,[fname ‘.driver’]) ;