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AMBER Archive (2006)
Subject: AMBER: Difference in results for TI calculation
From: brmeher_at_iitg.ernet.in
Date: Mon May 29 2006 - 08:53:41 CDT
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Dear Dr. Case,
I have read your
tutorial on thermodynamic integration (TI) and I was performing the
same calculation (Toluene---> nothing) in the tutorial given. But
unfortunately I got some differnt results as compared to the
tutorial.
First of all I created the prmtop.coul.vac
and prmcrd.coul.vac files from Xleap. Then I used these files to run
for a charging free energy calculation in vaccum and interestingly I
got almost same result of DV/DL (-0.4625) compared to tutorial DV/DL
( -0.4646). Then I created the prmtop.coul.wat and prmcrd.coul.wat
files for the charging free energy calculation in water. I used the
same input files as given in the tutorial for both equilibration and
production part with three different values of clambda (0.11270, 0.5
& 0.88729) and finally I got the more or less same results as
follows:
| clambda | DV/DL from my
calculation | DV/DL from tutorial |
| 0.11270 | 4.4439 +/- 2.1355 | 4.13 +/- 2.26 |
| 0.5 | 1.5345 +/- 1.7721 | 1.52 +/- 1.80 |
| 0.88729 | -0.3335 +/- 1.42 | -0.26 +/- 1.49 |
So my question is that if I am using the same input as per the tutorial and also the same prmtop & prmcrd files , why this difference in results is coming??
Secondly I also tried for the
Disappearing free energy calculation for both in vaccum and in
water. For vaccum part I got the DV/DL of -1.8537 . the
input for the vaccum is as follows:
Here is the input
file:
neutralize toluene
&cntrl
ntr=0,
nstlim
=100000, nscm=2000, ntave=10000,
ntx=5, irest=1, ntb=0,
ntpr=200,
ntp=0, taup=2.0,
dt=0.001,
nrespa=2,
ntt=1, temp0=300., tautp=2.0,
ntc=2,
ntf=2, tol=0.000001,
ntwr = 10000, ntwx=0,
icfe=1,
klambda=4,
cut=9.0,
&end
For disappearing
free energy calculation in water part I equilibrated the system for each
individual values of clambda and then run for the production for the same
value of clambda. I used five values of
clambda from 0.05, 0.15,0.25, 0.35 and 0.45 with klambda =4.
Here is my result which is far different from the results of your
tutorial.
| clambda | DV/DL from my
calculation |
| 0.05 | 13.2628 |
| 0.15 | 4.1287 |
| 0.25 | 1.3485 |
| 0.35 | 0.0854 |
| 0.45 | -0.0716 |
Here is an
input file for an example for clambda = 0.05
neutralize toluene
&cntrl
ntr=0,
nstlim
=100000, nscm=2000, ntave=5000,
ntx=7, irest=1, ntb=1, ntpr=100,
ntp=0, taup=2.0,
dt=0.001, nrespa=2,
ntt=0, temp0 = 300.,
tautp=2.0,
ntc=2, ntf=2, tol=0.000001,
ntwr = 10000,
ntwx=0,
icfe=1, clambda=0.05, klambda=4,
cut=9.0,
&end
I
am using the AMBER-8 version in a Linux system.
So can you
please point out the reason of differnce in results and also the
mistake.
With regards
Biswa Ranjan Meher
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- Next message: brmeher_at_iitg.ernet.in: "AMBER: Difference in results for TI calculation"
- Previous message: Vitor Manuel Sousa F?x: "AMBER: Thermodynamic integration using the amber 9"
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