Residuals of the static equations:
Equation number 1 : -10.4446
Equation number 2 : 3.8
Equation number 3 : 1.6955
Equation number 4 : Inf
Equation number 5 : 0.86647
Equation number 6 : -0.31565
Equation number 7 : 0
Equation number 8 : Inf
Equation number 9 : -0.95
Equation number 10 : 2.8183
STEADY: The Jacobian contains Inf or NaN. The problem arises from:
STEADY: Derivative of Equation 4 with respect to Variable y (initial value of y: 5000)
STEADY: Derivative of Equation 8 with respect to Variable y (initial value of y: 5000)
STEADY: The problem most often occurs, because a variable with
STEADY: exponent smaller than 1 has been initialized to 0. Taking the derivative
STEADY: and evaluating it at the steady state then results in a division by 0.
tech.mod (1.18 KB)
You cannot use 5000 as a starting value for log output as you will be using
in the model. Also
does not look like a correct equation. Why is there a division by g on the RHS?
Dear Professor Jpfeifer, Thanks for valuable suggestion and guide line. But I am unable to solve after correction. I try to solve for steady state value by using MATLAB under .m file. its looks very difficult. Is there any helpful reference for getting initial value? Please help for solving this model.
tech.mod (1.3 KB)
Your equations are still wrong.
has steady state
Please make sure the exp() is consistently used in a correct way.
Please help, Regarding this issues. Thanks in advanced.
“An infinite element was encountered when trying to solve equation(s) 3
with respect to the variable(s): pi.”
tech.mod (1.37 KB)
Again, make sure the exp()-substitution is consistent. In the first equation,
rn is not in exp(). In
exp(Ln((1+rn(+1))/(1+rn))) it is. For the start, I would make the model run without this type of substitution and only then add the exp()
Now I try to find out Steady state value by using “steady_state_model” command for the instant of the initial value. Plz help and thanks in Advanced.
tech.mod (1.7 KB)
I can only repeat myself. You did not head any of my previous advice. Note that a
block requires the full analytical steady state to be computed. It does not seem you did this. For example, z has steady state 0 in the entered AR-process. You have a value for c hard-coded, which also does not seem plausible given that you used an exp()-substitution that implies c is in logs.
Dear Professor Johannes Pfeifer, I follow your suggestions from the first. But I can not find out the solution. So I try to another way. Now I can understand that it’s very difficult. Plz see the attachment, I have corrected by your last suggestions.
techT.mod (1.37 KB)
See my very first comment on
The way you do the exp()-substitution is still not consistent
I have done. Is it ok?
techT.mod (1.36 KB)
This is leading nowhere. For the last time, please sit down and check every single equation on whether the exp()-substitution has been done consistently. The last equation for example has an
not in exp() while it is in exp() everywhere else. As I suggested in the beginning, you should try to solve the model first without any substitution and make sure it runs. That makes debugging much easier.
I rewrite this equation as follows
Please see the attachment. techT.mod
And Alternative file without substitution: Attachment: techW.mod
techW.mod (1.25 KB)
techT.mod (1.36 KB)
Please help regarding this matter. Thanks in advanced.
I don’t think you are listening. You are supposed to write down the file without the exp-substitution. In
techW.mod you did that, but you also deleted the
exp() required for the exogenous processes. Another issue seems to be that your model features an infinity of steady states. You may have the problem mentioned at An infinity of steady states with Taylor rules