Dear Mark,

Thanks for your reply. 
Before this test, I have performed several thermodynamics integration playing 
with the number of lambda points and also decoupling first coulomb and then VDW 
or decoupling both at the same time. All my tests gave the same results as long 
as I am not using tabulated form of potential. In all of them, I obtained 
74kj/mol as solvation free energy!

Now, my question is that, why using tabulated potential makes that much of a 
difference?
Do you mean that decoupling both Q and VDW can be a problem while using 
tabulated potential? In any case, I will lunch a simulation decoupling them 
separately and will update you.

I really appreciate your help 
Fahimeh


Hi,

Decoupling both q and vdw at the same time is not a good idea. You should
observe that at nearly-disappeared lambda, the error bars grow dramatically.

Mark




Hi everyone,

I am computing solvation free energy of molecule A in solvent B. I used two 
different procedure and I thought that the results should be comparable, 
however they are very different. 

Procedure 1: 
Thermodynamics integration using 20 lambda points. VDWtype was switch and I 
used PME coulombtype.  I turned off VDW and coulomb interactions both at the 
same time. The free energy of solvation was  74kj/mol

Procedure 2:
Thermodynamics integration using 20 lambda points. I introduce tabulated 
potential for the interaction between molecule A and molecule B. therefore the 
vdwtype and columbtype was set to User. I provided two files, called table.xvg 
and table_A_B.xvg which are identical for now as a test. These files contain, 
r, 1/r, 1/r**2, -1/r**6, -6/r**7, 1/r**12, 12/r**13.  The other parameters for 
VDW interactions are untouched. The result of this simulation however is very 
different ~ 50kj/mol ... 

I am expecting to receive not identical but something comparable ... could you 
please help me to understand this?

Many thanks in advance
-- 
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