NONMEM Users Network Archive

Hosted by Cognigen

Limit for model complexity in NONMEM?

From: Martin Fransson <marfr>
Date: Tue, 19 Dec 2006 09:32:37 +0100

Dear all,
 

I am a novice in NONMEM and am currently trying to fit data for a parent =
substance (P) and its two metabolites (Ma resp. Mb). As you can see =
below I use a nine compartment model with three compartments for each =
compound. The data (ln-transformed) is measured concentrations from =
infusion treatment and the total amount of data is approximately 20 =
(patients) x 10 (samples) x 3 (compounds), i.e., about 600 data records. =
Dose is applied to compartment P1. Since I am still trying to figure out =
the features of NONMEM (working with v. 6) perhaps you can help with me =
some questions I have.

 

1) Is it at all sensible to use nine compartment models in NONMEM? Is =
there an unspoken maximum complexity (in terms of no of parameters, no =
of compartments, "degree" of nonlinearity etc.) for the models one =
should use? As you can see there are already 10 unknown parameters with =
equally many random effects. If I a introduce nonlinearities such as =
MM-kinetics it is most likely that there will be even more parameters (I =
guess making the model nonlinear will also make things more difficult =
for the optimization algorithm). Investigating effects of covariates =
will also require additional parameters. (Some models I have tried did =
have up to 18 unknown parameters.)

 

2) Are there any obvious errors in the code? For instance when it comes =
to the aspect of modeling parent and metabolites at the same time?



3) It seems that it is almost impossible to get FOCE with interaction to =
work when using complex models, but if I assume CCV-error, perform =
ln-transformation of the data and use FO instead will it make any =
difference?

 

Notes:

1) I use Advan6 since I intend to make parts of the model nonlinear at =
later stage.

2) I do not capture any exceptions in the ERROR record in form of LOG(0) =
since I removed any zero valued data records.



All comments and critics are highly appreciated!

 

Best regards,

Martin





$PROBLEM Model 0 (linear),
$INPUT ID AMT=DOSE RATE TIME CP=DV CMT EVID MDV CO1 CO2 CO3 AGE BSA
$DATA DCPLog.txt
$SUBROUTINES ADVAN6 TOL=5
$MODEL COMP (P1) COMP (P2) COMP (P3)
       COMP (Ma1) COMP (Ma2) COMP (Ma3)
       COMP (Mb1) COMP (Mb2) COMP (Mb3)
$PK
V1=THETA(1)*EXP(ETA(1))
V2=THETA(2)*EXP(ETA(2))
V3=THETA(3)*EXP(ETA(3))
Q12=THETA(4)*EXP(ETA(4))
Q13=THETA(5)*EXP(ETA(5))

;Enzyme kinetics
kPMa=THETA(6)*EXP(ETA(6))
kPMb=THETA(7)*EXP(ETA(7))
kelP=THETA(8)*EXP(ETA(8))
kelMa=THETA(9)*EXP(ETA(9))
kelMb=THETA(10)*EXP(ETA(10))

$DES
x1=A(1)/V1
x2=A(2)/V2
x3=A(3)/V3
x4=A(4)/V1
x5=A(5)/V2
x6=A(6)/V3
x7=A(7)/V1
x8=A(8)/V2
x9=A(9)/V3

DADT(1)=-Q12*(x1-x2)-Q13*(x1-x3)
DADT(2)=Q12*(x1-x2)-V2*(kPMa+kPMb+kelP)*x2
DADT(3)=Q13*(x1-x3)
DADT(4)=-Q12*(x4-x5)-Q13*(x4-x6)
DADT(5)=Q12*(x4-x5)+V2*kPMa*x2-V2*kelMa*x5
DADT(6)=Q13*(x4-x6)
DADT(7)=-Q12*(x7-x8)-Q13*(x7-x9)
DADT(8)=Q12*(x7-x8)+V2*kPMb*x2-V2*kelMb*x8
DADT(9)=Q13*(x7-x9)

$ERROR CALLFL=0
IF (CMT.EQ.1) z=A(1)/V1
IF (CMT.EQ.4) z=A(4)/V1
IF (CMT.EQ.7) z=A(7)/V1
Y=LOG(z)+LOG(1+EPS(1))

$THETA
(3,5) ; V1
(1,3) ; V2
(30,70) ; V3
(0.01,1.5) ; Q12
(0.001,0.02) ; Q13
(0.0001,0.01)
(0.0001,0.01)
(0.0001,0.01)
(0.0001,0.01)
(0.0001,0.01)

$OMEGA .1,.1,.1,.1,.1,.1,.1,.1,.1,.1,
$SIGMA .15

$EST METHOD=0 MAXEVAL=9999 PRINT=1
$COV
;$TABLE TIME PRED
$SCAT CP VS TIME
$SCAT PRED VS TIME
$SCAT PRED VS CP UNIT
$SCAT RES VS TIME


==========================
======
Martin Fransson
Dept. of Computer and Information Science
Linköping University
581 83 LINKÖPING, SWEDEN
marfr
+46 13 281467
==========================
======
Received on Tue Dec 19 2006 - 03:32:37 EST

The NONMEM Users Network is maintained by ICON plc. Requests to subscribe to the network should be sent to: nmusers-request@iconplc.com.

Once subscribed, you may contribute to the discussion by emailing: nmusers@globomaxnm.com.