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Hi,everyone:

I have a question about the steady state and QZ decomposition of my linear model.
the result shows that:
STEADY-STATE RESULTS:

y 0
c 0
k 0
i 0
n 0
m 0.26738
b 0
w 0
rk 0
mc 0
rn 0
pai 0
g 0
a 0
??? Error using ==> print_info at 36
The generalized Schur (QZ) decomposition failed. For more information, see the documentation
for Lapack function dgges: info=13, n=11

Error in ==> check at 76
print_info(info, options.noprint);

Error in ==> code at 243
oo_.dr.eigval = check(M_,options_,oo_);

Error in ==> dynare at 120
evalin('base',fname) ;


I try to get the steady state value of"m",but it isn't 0.26738, I think it should be 0. and the The generalized Schur (QZ) decomposition is failed .Is there any error of my model?

I have confused by this queation for many weeks,If anybody can offer help, I'd really appreciate it.

First, focus on the steady state. Put

Code: Select all

resid(1);

before steady to see the that the problematic equations are 3 and 4. In 4, pai should be 0, but you initialize it to 1. When you correct this, you can see that equation 3 has a problem. The reason is the 1 in that equation. I guess the log-linearization is wrong.

Regarding the QZ: model_diagnostics says:

model_diagnostic: the Jacobian of the static model is singular
there is 1 colinear relationships between the variables and the equations
Colinear variables:
y
c
k
i
n
b
w
mc
rn
Colinear equations
6 7 8 10

The presence of a singularity problem typically indicates that there is one
redundant equation entered in the model block, while another non-redundant equation
is missing. The problem often derives from Walras Law.

jpfeifer wrote:First, focus on the steady state. Put

Code: Select all

resid(1);

before steady to see the that the problematic equations are 3 and 4. In 4, pai should be 0, but you initialize it to 1. When you correct this, you can see that equation 3 has a problem. The reason is the 1 in that equation. I guess the log-linearization is wrong.

Regarding the QZ: model_diagnostics says:

model_diagnostic: the Jacobian of the static model is singular
there is 1 colinear relationships between the variables and the equations
Colinear variables:
y
c
k
i
n
b
w
mc
rn
Colinear equations
6 7 8 10

The presence of a singularity problem typically indicates that there is one
redundant equation entered in the model block, while another non-redundant equation
is missing. The problem often derives from Walras Law.

Sorry for seeing the reply for so long time. But I have a question, what's the effect of "resid(1);"? And when shoud I use it? Thanks!

It displays the residuals of the static model equations. Put it before steady and after initival to see how good your starting values are.