The generalized Schur (QZ) decomposition failed

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.
dual.mod (3.38 KB)

says

[quote]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[/quote]


Look at equation 16:
[code]y=(y_S_ss)^rho/y^rho*y_S+(y_L_ss)^rho/y^rho*y_L;
[/code]
with you dividing by y, this equation clearly is not linear.

says

[quote]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[/quote]

Look at equation 16:

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.

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.

[quote=“jpfeifer”]model_diagnostics(M_,options_,oo_)
says

[quote]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[/quote]

Look at equation 16:

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.[/quote]

dual3.mod (3.25 KB)

Hi,

if you run again:

it says:

[quote]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

[/quote]

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

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.

Hi,thank you for you advice. I have correct the mistake but the program still fails to process.=-=

[quote=“kyri82”]Hi,

if you run again:

model_diagnostics(M_,options_,oo_)

it says:

[quote]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

[/quote]

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

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.[/quote]

dual3.mod (3.26 KB)

As Kyriacos correctly says

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.