RE: Question about mixture model of Ka

From: Xiaofeng Wang Date: July 24, 2012 technical Source: mail-archive.com
Hi Claire, I think the error comes from your first IF statement. See the changes below. Best, Xiaofeng IF (OCC == 1) THEN BOVKA=ETA(3) ENDIF IF (OCC == 2) THEN BOVKA=ETA(4) ENDIF IF (OCC == 3) THEN BOVKA=ETA(5) ENDIF IF (OCC == 4) THEN BOVKA=ETA(6) ENDIF IF (OCC == 5) THEN BOVKA=ETA(7) ENDIF IF (MIXNUM==1) THEN ; pop1 KA=THETA(3) ELSE KA=THETA(4)*EXP(ETA(8)+BOVKA); pop 2 ENDIF
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________________________________ From: [email protected] [[email protected]] on behalf of Xu, Claire [[email protected]] Sent: Tuesday, July 24, 2012 10:56 AM To: Mats Karlsson Cc: [email protected] Subject: Re: [NMusers] Question about mixture model of Ka Hi Mats and nmusers who may be interested in this topic, Thanks a lot for your suggestion first and it helps! The estimation for subpopulation with faster absorption is about 11 and for the other subpopulation is about 1, which are close to expectation. And I was able to improve the fitting for two individuals who showed fast absorption in all occasions. As you may remembered that I posted a series of questions related to absorption models recently. The PK data that I have is a four-way cross-over study with a single oral dose administered to 26 healthy subjects in each arm. I have about 15% subjects having fast absorption, 10% with double peak and about 75% with kind of normal absorption profile. The problem now becomes that there is a difference in the PK profiles from the same individual across different occasions, e.g.a same individual has all three different absorption patterns listed above. Therefore, I tried to incorporate IOV to my mixture model of Ka. I want to put IOV on the Ka for normal absorption profile (or is there any better way to resolve my issue?). Below is my code and the error message is that ENDIF structure is wrong. Anyone has any advice? Thanks a lot for all the great input! $PK CL=THETA(1)*EXP(ETA(1)) V=THETA(2)*EXP(ETA(2)) EST=MIXEST IF (MIXNUM==1) THEN ; pop1 KA=THETA(3) ELSE IF (OCC == 1) THEN BOVKA=ETA(3) ENDIF IF (OCC == 2) THEN BOVKA=ETA(4) ENDIF IF (OCC == 3) THEN BOVKA=ETA(5) ENDIF IF (OCC == 4) THEN BOVKA=ETA(6) ENDIF IF (OCC == 5) THEN BOVKA=ETA(7) ENDIF KA=THETA(4)*EXP(ETA(8)+BOVKA); pop 2 ENDIF S2=V/1000 ; convert concentration to ng/ml $MIX NSPOP=2 A=EXP(THETA(5)) DEN=1+A P(1)=1/DEN ;pop1 P(2)=A/DEN ;pop2 $ERROR DEL = 0.0001 IPRED = LOG(DEL) IF (F.GT.DEL) IPRED = LOG(F) W=THETA(6) IRES=IPRED-DV IWRES = IRES/W Y=IPRED+EPS(1)*W $THETA (0,50) ;CL (0,550); V (0,200);KA for pop 1 (0,1); KA for pop 2 (0,1); per (0,0.16);W $OMEGA BLOCK(2) 0.16 0.01 0.16 $OMEGA BLOCK(1) 0.16 ;OCC1 $OMEGA BLOCK(1) SAME; OCC2 $OMEGA BLOCK(1) SAME; OCC3 $OMEGA BLOCK(1) SAME; OCC4 $OMEGA BLOCK(1) SAME; OCC5 $OMEGA 0.16 On Mon, Jul 23, 2012 at 3:37 PM, Mats Karlsson <[email protected]<mailto:[email protected]>> wrote: Dear Claire, Through your constraint on THETA(5), you have restricted the mixture model to have MIXNUM=2 as the dominating (>=50%) subpopulation. I think this may result in your run ending up in a local minimum. I wouldn’t be surprised if THETA(5) is driven towards zero. If you want to keep this parameterization for P(1), maybe you should try switching initial est’s between THETA3 and THETA4. Best regards, Mats Mats Karlsson, PhD Professor of Pharmacometrics FIRST WORLD CONFERENCE ON PHARMACOMETRICS, 5-7 September 2012, Seoul http://www.go-wcop.org/) Dept of Pharmaceutical Biosciences Faculty of Pharmacy Uppsala University Box 591 75124 Uppsala Phone: +46 18 4714105<tel:%2B46%2018%204714105> Fax + 46 18 4714003<tel:%2B%2046%2018%204714003> From: [email protected]<mailto:[email protected]> [mailto:[email protected]<mailto:[email protected]>] On Behalf Of Xu, Claire Sent: 23 July 2012 20:45 To: [email protected]<mailto:[email protected]> Subject: [NMusers] Question about mixture model of Ka Hi Dear nmusers, I am trying to develop a pop PK model. In my data, there are two subpopulations, one subpopulation with Ka of about 9 and the other one with Ka of about 1. The percentage of population with higher Ka is about 15%. I assumed that it would be reasonable to utilize mixture model of Ka. But I cannot get the mixture model of Ka successfully run under NONMEM, because the percentage of the subpopulation can't be estimated. Can anyone help check my NONMEM code? Is there anything special that I should do for the input data profile? .I will very appreciateit it for your generous help. Thanks a lot! Claire $SUB ADVAN2 TRANS2 $PK CL=THETA(1)*EXP(ETA(1)) V=THETA(2)*EXP(ETA(2)) EST=MIXEST IF (MIXNUM==1) THEN ; pop1 KA=THETA(3)*EXP(ETA(3)) ELSE KA=THETA(4)*EXP(ETA(4)) ; pop2 ENDIF S2=V/1000 ; convert concentration to ng/ml $MIX NSPOP=2 A=EXP(THETA(5)) DEN=1+A P(1)=1/DEN ;pop1 P(2)=A/DEN ;pop2 $ERROR DEL = 0.0001 IPRED = LOG(DEL) IF (F.GT.DEL) IPRED = LOG(F) W=THETA(6) IRES=IPRED-DV IWRES = IRES/W Y=IPRED+EPS(1)*W $THETA (0,50) ;CL (0,550); V (0,1.5);KA for pop 1 (0,6); KA for pop 2 (0,1); per (0,0.16);W $OMEGA BLOCK(2) 0.16 0.01 0.16 $OMEGA 0.16 0.16 -- Xu, Claire Ph.D Candidate Division of Clinical Pharmacology, Wishard Hospital Indiana University School of Medicine 1001 West 10th Street, Myers W7122 Indianapolis, IN 46202 T - 317/7558242<tel:317%2F7558242> -- Xu, Claire Ph.D Candidate Division of Clinical Pharmacology, Wishard Hospital Indiana University School of Medicine 1001 West 10th Street, Myers W7122 Indianapolis, IN 46202 T - 317/7558242<tel:317%2F7558242>
Jul 23, 2012 Claire Xu Question about mixture model of Ka
Jul 23, 2012 Mats Karlsson RE: Question about mixture model of Ka
Jul 24, 2012 Claire Xu Re: Question about mixture model of Ka
Jul 24, 2012 Xiaofeng Wang RE: Question about mixture model of Ka