RE: calculation of time-to-nadir

From: Nick Holford Subject: RE:[NMusers] calculation of time-to-nadir Date: Tue, August 3, 2004 3:49 am Erik, Thanks very much for suggesting this solution to finding the time of maximum (or minimum) of a function. I thought it might be of interest to show the data, code and output for the kind of problem that Anthe asked about originally i.e. the time to nadir. This example implements a linear pharmacodynamic effect inhibiting the input to a turnover model so that the solution to the time of nadir is known analytically (as in your Cmax example). It uses the $SIMULATION option to simulate data from a template data file (nadir.dat). The output table file shows that your clever brute force method agrees quite closely with the analytical solution for the time of nadir. Note that the accuracy of the solution depends on the spacing of the times in the data set. When the data has only two times, 0 and 100 (time increment 100), then the predicted time of nadir is 54.884 compared with the analytical value of 54.931. Interestingly if the time increment is 1 unit then the brute force solution is somewhat worse (54.965) than with a time increment of 10 (54.933; see below). By trial and error I looked for the smallest value of TOL that would give a brute force prediction that is identical (54.931) to the analytical solution. Paradoxically I found it needed TOL=7 with a time increment of 100, TOL=8 with a time increment of 10 and TOL=9 with a time increment of 1. Nick $PROBLEM Nadir $INPUT TIME AMT CMT DV $DATA nadir.dat $SUBROUTINES ADVAN6 TOL=6 $MODEL COMP = CENTRAL COMP = EFFECT COMP = NADIR COMP = TMIN $PK K = THETA(1) KOUT = THETA(2) C0 = THETA(3) SLOPE = THETA(4) KIN=KOUT*C0 F2=C0 $DES DCP=A(1) DCE=A(2) DADT(1) = -K*DCP D2 = KIN*(1-SLOPE*DCP) - DCE*KOUT DADT(2) = D2 IF (D2.LT.0) THEN DADT(3) = D2 ; Y(3) is conc 'lost' from effect site DADT(4) = 1 ; Y(4) is elapsed time up to nadir ELSE DADT(3) = 0 DADT(4) = 0 ENDIF $ERROR CP = A(1) CE = A(2) NADIR = C0+A(3) TNAD = A(4) ;Theoretical time of nadir for limear PD model TMIN = (LOG(K)-LOG(KOUT))/(K-KOUT) Y = CE*(1+ERR(1)) $THETA 0.01 ; k 0.03 ; keq 100 ; C0 0.01 ; Slope $OMEGA 0.001 ; cverr $SIM (123456) SUBPROBLEMS=1 ONLYSIM $TABLE TIME CP CE NADIR TNAD TMIN NOPRINT ONEHEADER NOAPPEND FILE=nadir.fit nadir.dat #TIME AMT CMT DV 0 1 2 . 0 100 1 . 10 . . . 20 . . . 30 . . . 40 . . . 50 . . . 60 . . . 70 . . . 80 . . . 90 . . . 100 . . . nadir.fit TIME CP CE NADI TNAD TMIN 0 0 100 100 0 54.931 0 100 100 100 0 54.931 10 90.484 75.397 75.397 10 54.931 20 81.873 59.512 59.512 20 54.931 30 74.082 49.863 49.863 30 54.931 40 67.032 44.631 44.631 40 54.931 50 60.653 42.49 42.49 50 54.931 60 54.881 42.473 42.265 54.933 54.931 70 49.659 43.881 42.265 54.933 54.931 80 44.933 46.208 42.265 54.933 54.931 90 40.657 49.095 42.265 54.933 54.931 100 36.788 52.286 42.265 54.933 54.931 -- Nick Holford, Dept Pharmacology & Clinical Pharmacology University of Auckland, 85 Park Rd, Private Bag 92019, Auckland, New Zealand email:n.holford@auckland.ac.nz tel:+64(9)373-7599x86730 fax:373-7556 http://www.health.auckland.ac.nz/pharmacology/staff/nholford/