Re: stepwise zero order process
From: Mats Karlsson <Mats.Karlsson@biof.uu.se>
Subject: Re: stepwise zero order process
Date: Wed, 07 Oct 1998 10:41:06 +0200
Dear Ralph,
Below you'll find files for doing zero-order stepwise input into NONMEM. THe files are for single subject data. Previously we have used the same approach to do population deconvolution of cyclosporin (Biopharm. Drug Disp (15):75-85 1994.). I haven't diagnozed the problem you experienced but a general caution for using zero-order input in NONMEM is to be observant what really happens (which at least to me sometimes can be surprising). For example, the default for varying bioavailability is that it affects the duration rather than rate of the input.
Best regards,
Mats
$PROB FLEXIBLE INPUT SINGLE SUBJECT ANALYSIS
$INPUT ID AMT TIME DV FLAG EVID MDV RATE
$DATA BA2.dta NRECS=48 NOREWIND IGNORE=#
$SUBS ADVAN3 TRANS3
$PK
; THETA(5) = FRACTION ABSORBED DURING 0-.5 HOURS
; THETA(6) = FRACTION ABSORBED DURING .5-1 HOURS
; THETA(7) = FRACTION ABSORBED DURING 1-2 HOURS
; THETA(8) = FRACTION ABSORBED DURING 2-4 HOURS
; THETA(9) = FRACTION ABSORBED DURING 4-6 HOURS
; THETA(10)= FRACTION ABSORBED DURING 6-9 HOURS
; THETA(11)= FRACTION ABSORBED DURING 9-12 HOURS
; THETA(12)= FRACTION ABSORBED DURING 12-24 HOURS
; THETA(13)= FRACTION ABSORBED DURING 24-36 HOURS
; THETA(14)= FRACTION ABSORBED DURING 36-48 HOURS
; THETA(15)= FRACTION ABSORBED DURING 48-72 HOURS
; THETA(16)= FRACTION ABSORBED DURING 72-96 HOURS
; THETA(17)= FRACTION ABSORBED DURING 96-120 HOURS
; THETA(18)= FRACTION ABSORBED DURING 120-144 HOURS
;-----------------ABSORPTION MODEL-----------------------------
Q1=0 ; THESE LINES ARE USED TO
Q2=0 ; DESCRIBE THE FLEXIBLE
Q3=0 ; INPUT MODEL.
Q4=0 ; (UNTIL LINE 'D1=TOTD')
Q5=0
Q6=0
Q7=0
Q8=0
Q9=0
Q10=0
Q11=0
Q12=0
Q13=0
Q14=0
IF (TIME.LT..5) Q1=1
IF (TIME.LT.1.AND.TIME.GE..5) Q2=1
IF (TIME.LT.2.AND.TIME.GE.1) Q3=1
IF (TIME.LT.4.AND.TIME.GE.2) Q4=1
IF (TIME.LT.6.AND.TIME.GE.4) Q5=1
IF (TIME.LT.9.AND.TIME.GE.6) Q6=1
IF (TIME.LT.12.AND.TIME.GE.9) Q7=1
IF (TIME.LT.24.AND.TIME.GE.12) Q8=1
IF (TIME.LT.36.AND.TIME.GE.24) Q9=1
IF (TIME.LT.48.AND.TIME.GE.36) Q10=1
IF (TIME.LT.72.AND.TIME.GE.48) Q11=1
IF (TIME.LT.96.AND.TIME.GE.72) Q12=1
IF (TIME.LT.120.AND.TIME.GE.96) Q13=1
IF (TIME.GE.120) Q14=1
SVFR1 = THETA(6)
SVFR2 = THETA(6)
SVFR3 = THETA(7)
SVFR4 = THETA(8)
SVFR5 = THETA(9)
SVFR6 = THETA(10)
SVFR7 = THETA(11)
SVFR8 = THETA(12)
SVFR9 = THETA(13)
SVFR10 = THETA(14)
SVFR11 = THETA(15)
SVFR12 = THETA(16)
SVFR13 = THETA(17)
SVFR14 = THETA(18)
FR1=Q1*SVFR1
FR2=Q2*SVFR2
FR3=Q3*SVFR3
FR4=Q4*SVFR4
FR5=Q5*SVFR5
FR6=Q6*SVFR6
FR7=Q7*SVFR7
FR8=Q8*SVFR8
FR9=Q9*SVFR9
FR10=Q10*SVFR10
FR11=Q11*SVFR11
FR12=Q12*SVFR12
FR13=Q13*SVFR13
FR14=Q14*SVFR14
SC=FLAG-1
IV=2-FLAG
TOTF =FR1+FR2+FR3+FR4+FR5+FR6+FR7+FR8+FR9+FR10+FR11+FR12+FR13+FR14
TOTD1 =.5*Q1 + .5*Q2 +1*Q3 +2*Q4 +2*Q5 +3*Q6 +3*Q7
TOTD2 =12*Q8 + 12*Q9 +12*Q10 +24*Q11 + 24*Q12 + 24*Q13 + 24*Q14
F1 = TOTF*SC+IV
D1 = TOTD1+TOTD2
;--------------------DISPOSITION MODEL --------------------------
CL=THETA(1)
V =THETA(2)
Q =THETA(3)
VSS=THETA(4)
SC=V
; CL V Q VSS
$THETA (0,.9) (0,9) (0,1) (0,24)
;FRACS 1 2 3 4 5 6 7
$THETA -.1 (0,.002) (0,.02) (0,.07) (0,.1) (0,.05) (0,.05)
;FRACS 8 9 10 11 12 13 14
$THETA (0,.1) (0,.09) (0,.05) (0,.01) (0,.005) (0,.005) (0,.03)
; BASE
$THETA (0,1) -.5
$OMEGA BLOCK(2) .9 .01 .01
$ERROR
SC=FLAG-1
IV=2-FLAG
BASE=THETA(19)+(FLAG-1)*THETA(20)
SLOPE=THETA(5)*TIME/144*IV
CP=F+BASE+SLOPE
IPRED=CP
IRES=DV-IPRED
IWRES=IRES
Y = CP+ERR(1)+CP*ERR(2)
$EST PRINT=10 NOABORT MAXEVAL=9000
and the data file
1 0 0 0 1 0 1 0
1 160 0 0 1 1 1 0
1 0 0.08 17.2 1 0 0 0
1 0 0.25 16.2 1 0 0 0
1 0 0.5 15.8 1 0 0 0
1 0 1 13.1 1 0 0 0
1 0 2 11.8 1 0 0 0
1 0 4 8.7 1 0 0 0
1 0 6 6.7 1 0 0 0
1 0 9 5.7 1 0 0 0
1 0 12 4 1 0 0 0
1 0 24 2.3 1 0 0 0
1 0 36 3.1 1 0 0 0
1 0 48 1.6 1 0 0 0
1 0 72 1.4 1 0 0 0
1 0 96 1.6 1 0 0 0
1 0 120 1.1 1 0 0 0
1 0 144 1.1 1 0 0 0
1 160 0 0 2 4 1 -2
1 0 0 0.8 2 0 0 0
1 160 0.5 0 2 1 1 -2
1 0 0.5 0.7 2 0 0 0
1 160 1 0 2 1 1 -2
1 0 1 0.6 2 0 0 0
1 160 2 0 2 1 1 -2
1 0 2 1.1 2 0 0 0
1 160 4 0 2 1 1 -2
1 0 4 1.9 2 0 0 0
1 160 6 0 2 1 1 -2
1 0 6 2.9 2 0 0 0
1 160 9 0 2 1 1 -2
1 0 9 2.5 2 0 0 0
1 160 12 0 2 1 1 -2
1 0 12 2.8 2 0 0 0
1 160 24 0 2 1 1 -2
1 0 24 2.2 2 0 0 0
1 160 36 0 2 1 1 -2
1 0 36 2.1 2 0 0 0
1 160 48 0 2 1 1 -2
1 0 48 1.7 2 0 0 0
1 160 72 0 2 1 1 -2
1 0 72 1.2 2 0 0 0
1 160 96 0 2 1 1 -2
1 0 96 0.8 2 0 0 0
1 160 120 0 2 1 1 -2
1 0 120 0.8 2 0 0 0
1 160 144 0 2 1 1 -2
1 0 144 0.9 2 0 0 0
________________________________________________________
Mats Karlsson, PhD
Uppsala University
Div. of Biopharmaceutics and Pharmacokinetics
Dept. of Pharmacy/
Box 580, SE-751 23 Uppsala, Sweden
Internet: mats.karlsson@biof.uu.se
Phone: +46 18 471 41 05
Fax: +46 18 471 40 03