I am considering some monetary cycle model and one of my equation needs the variable value at steady state as a parameter, like
R=(Y/Yss)^rhoy*(Pi/Piss)^rhoPi
How can I use the value of steady state before solve it in dynare. 
Thanks
Further, in my model some dynamic equations doesn’t affect the steady state theoretically and they are only working when there is deviation from steady state, but when I try solving it in dynare, those equations have effect on the steady state value, like the interest rule I defined in the last posting.
write first the steady state model and solve. Then use these values to set Yss and Piss that should be parameters in the second model.
By the way, your interest rate rule looks weird because R=1 at the steady state. The natural rate of interest seems to be missing from your equation.
If equations that are written entirely in difference from the steady state affect the steady that is the sign that there is a mistake either in the steady state model or in the dynamic model
Best
Michel
Thanks, I made correction accordingly. Sorry, one more question. It seems I asked too much recently. I think Dynare count the static equations into the dynamic system, and produce some Infinite eigenvalues, Does it affect the further analysis? For example, infinite large variance for some variables. I confront such infinite variance problem but I don’t know whether it is caused by my model or didn’t use Dynare correctly.
No, the static equations and variables are eliminated before constructing the structural state representation.
Infinite eigenvalues are firmly grounded in the theory of generalized eigenvalues. They have no detrimental influence on the next steps. As far as Blanchard and Kahn conditions are concerned they have to be counted as explosive roots of modulus larger than one.
As far as the structure of your model is concerned, their existence mean that it could be possible thru substitutions to write the same model with less forward variables, but then the economic interpretation of the equations would be lost. Better, not to worry about infinite eigenvalues.
Best
Michel