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RE: COND LAPLACE LIKELIHOOD

From: Mats Karlsson <mats.karlsson>
Date: Wed, 5 Aug 2009 08:04:20 +0200

Dear Wang

Did you put proper boundaries on B2 & B3?

Best regards,
Mats

Mats Karlsson, PhD
Professor of Pharmacometrics
Dept of Pharmaceutical Biosciences
Uppsala University
Box 591
751 24 Uppsala Sweden
phone: +46 18 4714105
fax: +46 18 471 4003

-----Original Message-----
From: owner-nmusers
On Behalf Of wangx826
Sent: Tuesday, August 04, 2009 11:38 PM
To: Leonid Gibiansky
Cc: Samer Mouksassi; NMUSERS
Subject: Re: [NMusers] COND LAPLACE LIKELIHOOD

Leonid,

Thanks for your suggestion. I totally agree the part you indicated was
incorrect in my original control stream. But after I revised it, NONMEM
still stopped running and gave the same error message like " CONDITIONAL =

LIKELIHOOD SET TO NEGATIVE VALUE WITH INDIVIDUAL 2 (IN INDIVIDUAL RECORD =

ORDERING), DATA RECORD 13". Is there anything else wrong in my coding? I =

would appreciate pretty much if you or some one else could point it out =
for
me.

Thanks again,
Tianli

On Aug 4 2009, Leonid Gibiansky wrote:

>something is incorrect here, either model or comments:
> P0 = C0/(1+C0) ; Probability of Score=>0
> P1 = C1/(1+C1) ; Probability of Score=>1
> P2 = C2/(1+C2) ; Probability of Score=>2
>
>Probability of Score=>0 is always 1 (because score is 0 or positive
>number as I can see).
>
>My guess is that the model is incorrect, it should be (assuming that =
the
>drug increases probability of higher scores while natural state favors
>score 0)
>
> > A1 = B1 + DRUG
> > A2 = B1 - B2 + DRUG
> > A3 = B1 - B2 - B3 + DRUG
>
> > C1 = EXP(A1)
> > C2 = EXP(A2)
> > C3 = EXP(A3)
> > P1 = C1/(1+C1) ; Probability of Score=>1
> > P2 = C2/(1+C2) ; Probability of Score=>2
> > P3 = C3/(1+C3) ; Probability of Score=>3
> >
> > PR0 = 1-P1 ; Probability of Score=0
> > PR1 = P1-P2 ; Probability of Score=1
> > PR2 = P2-P3 ; Probability of Score=2
> > PR3 = P3 ; Probability of Score=3
>
> > IF (DV.EQ.0) Y=PR0
> > IF (DV.EQ.1) Y=PR1
> > IF (DV.EQ.2) Y=PR2
> > IF (DV.EQ.3) Y=PR3
>
> B2 and B3 should be positive
>
>--------------------------------------
>Leonid Gibiansky, Ph.D.
>President, QuantPharm LLC
>web: www.quantpharm.com
>e-mail: LGibiansky at quantpharm.com
>tel: (301) 767 5566
>
>
>
>
>wangx826
>> Hi Samer,
>> Thanks for your quick reply. Here is my control stream:
>>
>> $SUBS ADVAN6 TOL=6
>> $MODEL
>> COMP=(ABSO)
>> COMP=(CENT)
>> COMP=(PERI)
>> COMP=(EFFECT)
>> $PK
>> CL =ICL*24
>> V2 = IVC
>> Q=...
>> K23=Q/V2
>> V3=
>> K32=Q/V3
>> K=CL/V2
>> KA=...
>> KE0 = THETA(6)
>> B1 = THETA(1)
>> B2 = THETA(2)
>> B3 = THETA(3)
>> EMAX = THETA(4)
>> EC50 = THETA(5)*EXP(ETA(1))
>>
>> $DES
>> DADT(1)=-A(1)*KA
>> DADT(2)=A(1)*KA-A(2)*K23+A(3)*K32-A(2)*K
>> DADT(3)=A(2)*K23-A(3)*K32
>> DADT(4)=KE0*(A(2)/V2-A(4))
>>
>> $ERROR
>> CE=A(4)
>> DRUG = EMAX*CE/(EC50+CE)
>> A0 = B1 + DRUG
>> A1 = B1 + B2 + DRUG
>> A2 = B1 + B2 + B3 + DRUG
>> C0 = EXP(A0)
>> C1 = EXP(A1)
>> C2 = EXP(A2)
>> P0 = C0/(1+C0) ; Probability of Score=>0
>> P1 = C1/(1+C1) ; Probability of Score=>1
>> P2 = C2/(1+C2) ; Probability of Score=>2
>>
>> PR0 = P0 ; Probability of Score=0
>> PR1 = P1-P0 ; Probability of Score=1
>> PR2 = P2-P1 ; Probability of Score=2
>> PR3 = 1-P2 ; Probability of Score=3
>> IF (DV.EQ.0) Y=PR0
>> IF (DV.EQ.1) Y=PR1
>> IF (DV.EQ.2) Y=PR2
>> IF (DV.EQ.3) Y=PR3
>>
>> $THETA (-20 -6.3) ; THETA1 B1
>> $THETA (-10 -0.3) ; THETA2 B2
>> $THETA (-10 2) ; THETA3 B3
>> $THETA (0 5) ; THETA4 EMAX
>> $THETA (0 50) ; THETA5 EC50
>> $THETA (0 1) ; THETA6 KEO
>>
>> $OMEGA 2
>>
>> $ESTIMATION MAXEVAL=9999 PRINT=5 METHOD=COND LAPLACIAN =
LIKELIHOOD
>> NOABORT MSFO=MSF1
>> Here I include the dataset for first two subjects. Since it is a
>> simultaneous PKPD link model, DV in the data set as follows is
>> categorical PD data. The dose was given daily and DV was recorded =
daily
>> as well.
>> #SUBJ TIME AMT EVID II ADDL PD KA CL V2 =
MDV
>> 1 0 . 0 . . 0 9.96 4.12 63.57 0
>> 1 0 1 1 1 20 . 9.96 4.12 63.57 1
>> 1 1 . 0 . . 0 9.96 4.12 63.57 0
>> 1 2 . 0 . . 0 9.96 4.12 63.57 0
>> 1 3 . 0 . . 0 9.96 4.12 63.57 0
>> 1 4 . 0 . . 0 9.96 4.12 63.57 0
>> 1 5 . 0 . . 0 9.96 4.12 63.57 0
>> 1 6 . 0 . . 0 9.96 4.12 63.57 0
>> 1 7 . 0 . . 0 9.96 4.12 63.57 0
>> 1 8 . 0 . . 0 9.96 4.12 63.57 0
>> 1 9 . 0 . . 0 9.96 4.12 63.57 0
>> 1 10 . 0 . . 0 9.96 4.12 63.57 0
>> 1 11 . 0 . . 0 9.96 4.12 63.57 0
>> 1 12 . 0 . . 0 9.96 4.12 63.57 0
>> 1 13 . 0 . . 0 9.96 4.12 63.57 0
>> 1 14 . 0 . . 0 9.96 4.12 63.57 0
>> 1 15 . 0 . . 0 9.96 4.12 63.57 0
>> 1 16 . 0 . . 0 9.96 4.12 63.57 0
>> 1 17 . 0 . . 0 9.96 4.12 63.57 0
>> 1 18 . 0 . . 0 9.96 4.12 63.57 0
>> 1 19 . 0 . . 0 9.96 4.12 63.57 0
>> 1 20 . 0 . . 0 9.96 4.12 63.57 0
>> 1 21 . 0 . . 0 9.96 4.12 63.57 0
>> 1 22 . 0 . . 0 9.96 4.12 63.57 0
>> 1 23 . 0 . . 0 9.96 4.12 63.57 0
>> 1 24 . 0 . . 0 9.96 4.12 63.57 0
>> 1 25 . 0 . . 0 9.96 4.12 63.57 0
>> 1 26 . 0 . . 0 9.96 4.12 63.57 0
>> 1 27 . 0 . . 0 9.96 4.12 63.57 0
>> 2 0 . 0 . . 0 1.52 4.68 66.91 0
>> 2 0 2 1 1 20 . 1.52 4.68 66.91 1
>> 2 1 . 0 . . 0 1.52 4.68 66.91 0
>> 2 2 . 0 . . 0 1.52 4.68 66.91 0
>> 2 3 . 0 . . 0 1.52 4.68 66.91 0
>> 2 4 . 0 . . 0 1.52 4.68 66.91 0
>> 2 5 . 0 . . 0 1.52 4.68 66.91 0
>> 2 6 . 0 . . 0 1.52 4.68 66.91 0
>> 2 7 . 0 . . 0 1.52 4.68 66.91 0
>> 2 8 . 0 . . 0 1.52 4.68 66.91 0
>> 2 9 . 0 . . 0 1.52 4.68 66.91 0
>> 2 10 . 0 . . 0 1.52 4.68 66.91 0
>> 2 11 . 0 . . 0 1.52 4.68 66.91 0
>> 2 12 . 0 . . 1 1.52 4.68 66.91 0
>> 2 13 . 0 . . 1 1.52 4.68 66.91 0
>> 2 14 . 0 . . 1 1.52 4.68 66.91 0
>> 2 15 . 0 . . 1 1.52 4.68 66.91 0
>> 2 16 . 0 . . 1 1.52 4.68 66.91 0
>> 2 17 . 0 . . 1 1.52 4.68 66.91 0
>> 2 18 . 0 . . 1 1.52 4.68 66.91 0
>> 2 19 . 0 . . 1 1.52 4.68 66.91 0
>> 2 20 . 0 . . 1 1.52 4.68 66.91 0
>> 2 21 . 0 . . 3 1.52 4.68 66.91 0
>> 2 22 . 0 . . 3 1.52 4.68 66.91 0
>> 2 23 . 0 . . 3 1.52 4.68 66.91 0
>> 2 24 . 0 . . 3 1.52 4.68 66.91 0
>> 2 25 . 0 . . 3 1.52 4.68 66.91 0
>> 2 26 . 0 . . 3 1.52 4.68 66.91 0
>> 2 27 . 0 . . 3 1.52 4.68 66.91 0
>> 2 28 . 0 . . 2 1.52 4.68 66.91 0
>>
>> Thanks,
>> Tianli
>>
>>
>> =
*************************************************************************=
*****
>>
>> Tianli Wang
>> University of Minnesota,
>> Department of Pharmaceutics
>>
>> On Aug 4 2009, Samer Mouksassi wrote:
>>
>>> Can you include your control (or at least your $EST bloc)
>>> And the proportional odds PD data. The likelihood may go wild if =
some
>>> categories are very rare or non esistent. You need to gard against =
over
>>> or underflow.
>>>
>>> -----Original Message-----
>>> From: owner-nmusers
[mailto:owner-nmusers
>>> On Behalf Of wangx826
>>> Sent: 2009-08-04 13:38
>>> To: NMUSERS
>>> Subject: [NMusers] COND LAPLACE LIKELIHOOD
>>>
>>> Dear nmusers,
>>>
>>> Has anybody seen the following error message from NONMEM:
>>> "CONDITIONAL LIKELIHOOD SET TO NEGATIVE VALUE
>>> WITH INDIVIDUAL 2 (IN INDIVIDUAL RECORD ORDERING), DATA RECORD =
13"?
>>> What does it mean?
>>>
>>> I am trying to use conditional lapalacian likelihood method to =
generate
>>> a proportional odds model dealing with categorical PD data. Here is =
a
>>> little piece of my data set:
>>> SUBJ TIME AMT EVID II ADDL DV CL V
>>> MDV
>>> 2 0 . 0 . . 0 4.12 63
>>> 0
>>> 2 0 1 1 1 20 . 4.12 63
>>> 1
>>> 2 0.005 . 2 . . . 4.12 63
>>> 1
>>> 2 0.01 . 2 . . . 4.12 63
>>> 1
>>> .
>>> .
>>> .
>>> 2 12 . 0 . . 0 4.12 63
>>> 0
>>> 2 13 . 0 . . 1 4.12 63
>>> 0
>>> .
>>> .
>>> .
>>> But I don't think the problem exists in the data set because I tried =

>>> deleting the rows indicated in the error message, but the exactly =
same
>>> message still came out and NONMEM stopped running. I have no idea =
what's
>>>
>>> the problem for my model.
>>>
>>> Thanks in advance,
>>>
>>> Tianli
>>>
>>
>>
>
Received on Wed Aug 05 2009 - 02:04:20 EDT

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