RE: double peak absoprtions

Dear Virginie, In NONMEM you do not write out bioavailability parameters like F1 and F2 in your differential equations (at least not if you use the reserved variables like F1 etc.). Furthermore to have two parallel absorption paths with different lag-times you need 2 absorption compartments and you do not write out ALAG in the differential equations. For this approach to work, your dataset needs to contain two dose rows for each dose entering the entire dose into both compartment 1 and 2 (see dataset example in the bottom of this email). Below we have written an example of something you could try out. We have assumed that you are modeling only oral data and hence are estimating clearance and volume over F (CL/F and V/F). The code is written so that using differential equations (ADVAN 6, 8 or 13) can be avoided in favor of using the faster general linear solution (ADVAN5 or 7). More about this can be found in the NONMEM manual. Having random effects on change point parameters like lag-times are known to be computationally problematic with the FOCE estimation method. Alternatives to avoid this problem could be to use a chain of transit compartments for one of the absorption routs (the delayed one) or to apply a time dependent "sigmoidal Emax like" step function to KA2. ;------------------------------------------------------- TVCL=THETA(1) CL=TVCL*EXP(ETA(1)) TV=THETA(2) V=TV*EXP(ETA(2)) TVQ=THETA(3) Q=TVQ*EXP(ETA(3)) TV2=THETA(4) V2=TV2*EXP(ETA(4)) ALAG1 = THETA(5)*EXP(ETA(5)) ; Lag-time for fast absorption path (perhaps remove ETA) ALAG2 = THETA(6)*EXP(ETA(6))+ALAG1 ; Lag-time for slow absorption path LF1 = LOG(THETA(7)/(1-THETA(7))) ; Logit transform for F1 F1 = EXP(LF1+ETA(7))/(1+EXP(LF1+ETA(7)) ; Fraction of dose absorbed via the fast absorption path F2 = 1-F1 ; Fraction of dose absorbed via slow absorption path KA1=THETA(8)*EXP(ETA(8)) ; First order rate of absorption for fast absorption path KA2=THETA(9)*EXP(ETA(9)) ; First order rate of absorption for slow absorption path K1T3 = KA1 K2T3 = KA2 K3T0 = CL/V K3T4 = Q/V K4T3 = Q/V2 $DES ; $DES block can be removed if ADVAN5 is used DADT(1) = -K1T3*A(1) DADT(2) = -K2T3*A(2) DADT(3) = K1T3*A(1)+K2T3*A(2)+K4T3*A(4)-K3T4*A(3)-K3T0*A(3) DADT(4) = K3T4*A(3)-K4T3*A(4) ;------------------------------------------------------- Example of input data: ID TIME DV AMT EVID CMT 1 0 0 100 4 1 1 0 0 100 1 2 1 1 1.1 0 0 3 1 2 4.5 0 0 3 1 3 3.2 0 0 3 2 0 0 200 4 1 2 0 0 200 1 2 2 1 4.1 0 0 3 2 2 3.5 0 0 3 2 3 8.1 0 0 3 ;------------------------------------------------------- Sincerely, David Khan and Martin Bergstrand, Pharmacometrics Research Group, Department of Pharmaceutical Biosciences, Uppsala University