Limit for model complexity in NONMEM?

1 messages 1 people Latest: Dec 19, 2006

Limit for model complexity in NONMEM?

From: Martin Fransson Date: December 19, 2006 technical
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 MAXEVAL99 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 Linkping University 581 83 LINKPING, SWEDEN marfr +46 13 281467 ============= ===