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The generalized Schur (QZ) decomposition failed

PostPosted: Wed May 21, 2014 3:30 pm
by adi199283
Dear all. I am trying to run a mod file. But I just fail and the error is The generalized Schur (QZ) decomposition failed.
Does anybody has idea of how to solve it .
Many thanks.

Re: The generalized Schur (QZ) decomposition failed

PostPosted: Wed May 21, 2014 3:43 pm
by jpfeifer
Code: Select all
model_diagnostics(M_,options_,oo_)

says
STEADY: numerical initial values or parameters incompatible with the following equations
16

Check whether your model in truly linear

STEADY: The Jacobian contains Inf or NaN. The problem arises from:

STEADY: Derivative of Equation 16 with respect to Variable y_L (initial value of y_L: 0)
STEADY: Derivative of Equation 16 with respect to Variable y_S (initial value of y_S: 0)
STEADY: Derivative of Equation 16 with respect to Variable y (initial value of y: 0)
STEADY: Derivative of Equation 19 with respect to Variable ksi (initial value of ksi: 0)
STEADY: Check whether your model in truly linear\n
MODEL_DIAGNOSTICS: The steady state cannot be computed
MODEL_DIAGNOSTICS: Steady state contains NaNs

Look at equation 16:
Code: Select all
y=(y_S_ss)^rho/y^rho*y_S+(y_L_ss)^rho/y^rho*y_L;

with you dividing by y, this equation clearly is not linear.

Re: The generalized Schur (QZ) decomposition failed

PostPosted: Fri May 23, 2014 12:44 am
by adi199283
Thank you for you advice. But after modifying the code, the new problem is Blanchard Kahn conditions are not satisfied: indeterminacy. I have tried various paremeter but I still fail to overcome this problem.

jpfeifer wrote:
Code: Select all
model_diagnostics(M_,options_,oo_)

says
STEADY: numerical initial values or parameters incompatible with the following equations
16

Check whether your model in truly linear

STEADY: The Jacobian contains Inf or NaN. The problem arises from:

STEADY: Derivative of Equation 16 with respect to Variable y_L (initial value of y_L: 0)
STEADY: Derivative of Equation 16 with respect to Variable y_S (initial value of y_S: 0)
STEADY: Derivative of Equation 16 with respect to Variable y (initial value of y: 0)
STEADY: Derivative of Equation 19 with respect to Variable ksi (initial value of ksi: 0)
STEADY: Check whether your model in truly linear\n
MODEL_DIAGNOSTICS: The steady state cannot be computed
MODEL_DIAGNOSTICS: Steady state contains NaNs

Look at equation 16:
Code: Select all
y=(y_S_ss)^rho/y^rho*y_S+(y_L_ss)^rho/y^rho*y_L;

with you dividing by y, this equation clearly is not linear.

Re: The generalized Schur (QZ) decomposition failed

PostPosted: Fri May 23, 2014 8:45 am
by kyri82
Hi,

if you run again:

Code: Select all
model_diagnostics(M_,options_,oo_)


it says:

model_diagnostic: the Jacobian of the static model is singular
there is 1 colinear relationships between the variables and the equations
Colinear variables:
ksi_L
ksi_S
Colinear equations
Columns 1 through 13

1 2 3 4 5 6 7 8 9 10 11 12 13

Columns 14 through 26

14 15 16 17 18 20 21 23 24 25 27 28 29

Columns 27 through 29

30 31 32


Variables ksi_S and ksi_L are linearly dependent and the system is not invertible. The equation:

Code: Select all
y=(ksi_S_ss)*ksi_S+(ksi_S_ss)*ksi_L;


cannot simultaneously solve for both. Make sure you set up the problem correctly and that you solve it correctly. If you have colinearities it means that you have at least one equation that you don't need, that is the same as another one in the model. Of course, if you remove it it will also mean that you are missing one since you need as many equations as variables. That's why my guess is that you did not set up the model correctly...

K.

Re: The generalized Schur (QZ) decomposition failed

PostPosted: Fri May 23, 2014 9:58 am
by adi199283
Hiļ¼Œthank you for you advice. I have correct the mistake but the program still fails to process.=-=
kyri82 wrote:Hi,

if you run again:

Code: Select all
model_diagnostics(M_,options_,oo_)


it says:

model_diagnostic: the Jacobian of the static model is singular
there is 1 colinear relationships between the variables and the equations
Colinear variables:
ksi_L
ksi_S
Colinear equations
Columns 1 through 13

1 2 3 4 5 6 7 8 9 10 11 12 13

Columns 14 through 26

14 15 16 17 18 20 21 23 24 25 27 28 29

Columns 27 through 29

30 31 32


Variables ksi_S and ksi_L are linearly dependent and the system is not invertible. The equation:

Code: Select all
y=(ksi_S_ss)*ksi_S+(ksi_S_ss)*ksi_L;


cannot simultaneously solve for both. Make sure you set up the problem correctly and that you solve it correctly. If you have colinearities it means that you have at least one equation that you don't need, that is the same as another one in the model. Of course, if you remove it it will also mean that you are missing one since you need as many equations as variables. That's why my guess is that you did not set up the model correctly...

K.

Re: The generalized Schur (QZ) decomposition failed

PostPosted: Mon May 26, 2014 1:39 pm
by jpfeifer
As Kyriacos correctly says
Code: Select all
ksi=(ksi_S_ss)*ksi_S+(ksi_S_ss)*ksi_L;

yields the indeterminacy. Given the solution for ksi determined by the other equations, there are infinitely many combinations of ksi_S and ksi_L making the equation hold. Unless there are other equations determining how to split ksi into ksi_S and ksi_L there is no way to get around this indeterminacy problem.