Re: Problems with PD Modelling

From: Xin yi Date: February 12, 2011 technical Source: mail-archive.com
Hi thanks for your reply. Compartment 4 is already defined as my observation compartment, so it suffices to put IPRED=F? Xin Yi
Quoted reply history
On Sat, Feb 12, 2011 at 12:28 AM, Hong Lu <[email protected]> wrote: > Hi Xin yi, > > You may try defining compartment 4 is your observation compartment. > **And then in $ERROR module, put IPRED=A(4) or IPRED=F. > > Hong Lu > > > > > > On Fri, Feb 11, 2011 at 10:35 AM, xin yi <[email protected]> wrote: > >> Hi all, >> >> I'm new to nonmem and would be grateful for some pointers and help. I'm >> trying to do a pk-pd sequential modelling. I have modelled the PK data but >> I've encountered problems with PD. My PD model is a indirect inhibition >> response model. I have tried to change my initial estimates reduce the >> number of parameters in the model, but nothing seem to get me a good >> convergence. At different times, NONMEM gives me different error messages >> such as "parameter estimates near its boundary", "Minimisation successful. >> However problems occured with the minimization"(blank entries for omega or >> sigma estimates in the results file) *** or the "r matrix is >> algorithmically singular and non-positive definite". I have a few questions: >> >> 1) Are there any errors in the way I input my control and data file? >> 2) Under $ERROR in the control file, did I define it correctly with >> EFF=A(4), Y**=EFF+EFF*ERR(1)+ERR(2)*? Or should it be IPRED=F, >> Y=F+F*ERR(1)+ERR(2). >> 3) Why are they no estimates for sigma and omega in the results file? I >> have been constantly changing my initial estimates for omega and sigma but >> they always give me nil results. >> >> I appreciate any help on this matter. Thanks! >> >> Regards, >> X.Y. Ng >> * >> This is an example of my control file: >> >> $PROB RUN# pd_3_advan6 >> $INPUT ID TIME DV AMT CMT ADDL II MDV V2I V3I QI CLI KAI >> >> $DATA FINAL2.2.CSV IGNORE=C >> $SUBROUTINES ADVAN6 TRANS1 TOL=3 >> $MODEL >> COMP=DEPOT >> COMP=CENTRAL >> COMP=PERIPH >> COMP=EFFECT ;$MODEL defines the no of compartments in the model >> >> $PK >> >> V2=V2I >> V3=V3I >> Q=QI >> CL=CLI >> KA=KAI >> S2=V2 >> S3=V3 >> KIN=THETA(1)*EXP(ETA(1)) >> KOUT=THETA(2)*EXP(ETA(2)) >> IC50=THETA(3)*EXP(ETA(3)) >> F4=KIN/KOUT >> >> $DES >> DADT(1)=-KA*A(1) >> DADT(2)=KA*A(1)-Q/V2*A(2)+Q/V3*A(3)-CL/V2*A(2) >> DADT(3)=-Q/V3*A(3)+Q/V2*A(2) >> INH =A(2)/(IC50+A(2)) >> DADT(4)=KIN*(1-INH)-KOUT*A(4) >> >> $ERROR >> CP2=A(2)/S2 >> CP3=A(3)/S3 >> ;IPRED=F >> EFF=A(4) >> Y=EFF+EFF*ERR(1)+ERR(2) >> >> $THETA (0,0.281) ;POPKin >> $THETA (0,0.003) ;POPkout >> $THETA (0,2) ;POPIC50 >> >> $OMEGA 0.003 ;BSV Kin >> $OMEGA 0.003 ;BSV Kout >> $OMEGA 0.003 ;BSV IC50 >> >> $SIGMA 0.01 ;ERRCCV >> $SIGMA 0.0015 ;ERRADD >> >> $ESTIMATION METHOD=1 INTERACTION NOABORT MAXEVAL=9990 PRINT=10 POSTHOC >> $COVARIANCE >> $TABLE ID TIME DV AMT CMT NOPRINT ONEHEADER FILE=pd_3_advan6.TAB >> >> and an example of my data: >> >> * CID TIME DV AMT CMT ADDL II MDV V2I V3I QI CLI KAI 101 0 0 100 1 2 8 >> 1 44.55 11.78 1.07 3.37 0.62 101 0 0 1 4 0 0 1 44.55 11.78 1.07 3.37 >> 0.62 101 3 243 0 4 0 0 0 44.55 11.78 1.07 3.37 0.62 101 7 293 0 4 0 0 0 >> 44.55 11.78 1.07 3.37 0.62 101 11 261 0 4 0 0 0 44.55 11.78 1.07 3.37 >> 0.62 101 15 260 0 4 0 0 0 44.55 11.78 1.07 3.37 0.62 101 19 277 0 4 0 0 >> 0 44.55 11.78 1.07 3.37 0.62 101 23 290 0 4 0 0 0 44.55 11.78 1.07 3.37 >> 0.62 101 35 233 0 4 0 0 0 44.55 11.78 1.07 3.37 0.62 101 39 271 0 4 0 0 >> 0 44.55 11.78 1.07 3.37 0.62 101 43 274 0 4 0 0 0 44.55 11.78 1.07 3.37 >> 0.62 101 47 276 0 4 0 0 0 44.55 11.78 1.07 3.37 0.62 * >> * > > >
Feb 11, 2011 Xin yi Problems with PD Modelling
Feb 11, 2011 Hong Lu Re: Problems with PD Modelling
Feb 12, 2011 Xin yi Re: Problems with PD Modelling
Feb 12, 2011 Leonid Gibiansky Re: Problems with PD Modelling
Feb 13, 2011 Xin yi Re: Problems with PD Modelling
Feb 13, 2011 Leonid Gibiansky Re: Problems with PD Modelling