From LIAO@cosmos.prius.jnj.com Tue Apr 16 09:17:38 1996
Subject: Exp function MATH overflow

Hi dear NONMEM users:

I got this error msg "Exp function MATH overflow" when I used ADVAN2 for a 1-compartment open medel with first-order input.

I believe this might be due to the fact that I have PK sampling time up to 120 DAYS. I tried to use \$PRED to define the same model and try to use a conditional statement to set the exp function to

zero when Ka*T get >175. However, I got an error msg saying I can not use conditional statement involving the time. Any suggestions will be greatly appreciated. My \$PREP is listed below.
___________________________________________________________
\$PRED
CL =THETA(1)*EXP(ETA(1))
V2 =THETA(2)*EXP(ETA(2))
KA =THETA(3)*EXP(ETA(3))
K=CL/V2
D1=AMT/V2
A1=KA*D1/(KA-K)
EINA = 0.
T2=K*TIME
IF (-T2 .GT. -175) EINA=DEXP(-K*TIME)
EINB = 0.
T2=KA*TIME
IF (-T2 .GT. -175) EINB=DEXP(-KA*TIME)
F= A1*(EINA - EINB)
YCAL=F
Y=F*(1 + ERR(1))

****

From pascal Tue Apr 16 10:10:21 1996
Subject: Re: Exp function MATH overflow

The error msg you got for "not using conditional statement involving the time" was produced by NM-TRAN, who checks any coding that could put it into trouble for writing the symbolic differentiation of your population model. And this what happens in this case since you introduce a discontinuity in functions involving ETAs (either K*TIME or KA*TIME). You can off course fool NM-TRAN by using verbatim code in your \$PRED, since verbatim code is not checked by NM-TRAN. But I would not advocate such solution since in this case your full model (integrating the derivatives) would be definitively wrong.

A correct solution for your case would be to use the EXIT statement together with NOABORT option (thanks to Lewis Sheiner for indicating me this "trick") (see chap IV G, NM-TRAN Guide).

If we assume that PK sampling time up to 120 DAYS makes sense, here is one possible implementation of this mechanism using your \$PRED (allthough this could work with ADVAN2):

\$PRED
CL =THETA(1)*EXP(ETA(1))
V2 =THETA(2)*EXP(ETA(2))
KA =THETA(3)*EXP(ETA(3))
K=CL/V2
D1=AMT/V2
A1=KA*D1/(KA-K)
;
; 0.061 corresponds to 175/Tmax with Tmax = 120*24 = 2880 hours
;
IF (K.GT.0.061.OR.KA.GT.0.061) EXIT
EINA=DEXP(-K*TIME)
EINB=DEXP(-KA*TIME)
F= A1*(EINA - EINB)
YCAL=F
Y=F*(1 + ERR(1))
\$ESTIMATION ..... NOABORT

With this code, you simply refuse certain parameter values, which NONMEM minimization algorithm will try to avoid in next searches.

Good luck,
Pascal Girard
NONMEM Project Group

****

From alison Wed Apr 17 10:38:50 1996
Subject: EXP function Math Overflow

Sam Liao sent a question about error msg "Exp function MATH overflow"

I don't see how an overflow can occur with this model, if the thetas are constrained to be positive:

\$PRED
CL =THETA(1)*EXP(ETA(1))
V2 =THETA(2)*EXP(ETA(2))
KA =THETA(3)*EXP(ETA(3))
K=CL/V2
D1=AMT/V2
A1=KA*D1/(KA-K)
EINA = 0.
T2=K*TIME
IF (-T2 .GT. -175) EINA=DEXP(-K*TIME)
EINB = 0.
T2=KA*TIME
IF (-T2 .GT. -175) EINB=DEXP(-KA*TIME)
F= A1*(EINA - EINB)

What is more likely is an underflow, with large KA and large TIME.

We say in Users Guide III:

"It is probably safe to ignore floating point underflow, provided that the resulting value is set to zero; this is the default action specified by the IEEE 754 standard for floating point arithmetic."

Evidently, some Fortran systems do not allow underflow to zero. This can cause difficulties in many areas of the NONMEM system. Fortunaely, underflows in abbreviated code are fairly easy to avoid.

Pascal Girard suggested using the EXIT and NOABORT features of NONMEM. He says:

"With this code, you simply refuse certain parameter values, which NONMEM minimization algorithm will try to avoid in next searches."

This will certainly work, but has the disadvantage that it causes NONMEM to avoid large values of KA and K that might well be perfectly legitimate biologically. E.g., some subjects may in fact have such fast clearance that the concentration of drug becomes negligible relatively quickly.

I think it better to not use EXIT in this situation, but rather to allow EINA and/or EINB to become 0 in a manner acceptale to NM-TRAN. There is indeed a discontinuity in their values, but I think this less harmful than to arbitrarily rule out large values of certain thetas just because a given computer cannot handle very small floating point values.

Here's what I would suggest. (I'll use Sam's original test for KA*TIME, rather than the modification by Pascal).

\$PRED
CL =THETA(1)*EXP(ETA(1))
V2 =THETA(2)*EXP(ETA(2))
KA =THETA(3)*EXP(ETA(3))
K=CL/V2
D1=AMT/V2
A1=KA*D1/(KA-K)
T2=K*TIME
IF (-T2 .GT. -175) THEN
QA=1
DA=0
ELSE
QA=0
DA=175
ENDIF
EINA=QA*DEXP(-K*TIME+DA)
EINB = 0.
T2=KA*TIME
IF (-T2 .GT. -175) THEN
QB=1
DB=0
ELSE
QB=0
DB=175
ENDIF
EINB=QB*DEXP(-KA*TIME+DB)
F= A1*(EINA - EINB)
YCAL=F
Y=F*(1 + ERR(1))

The effect of DA and DB is to protect the exponential from underflowing for "bad" values of -KA*TIME, and the effect of QA and QB is to set the result to 0 afterwards.

QA,QB,DA,DB have no effect on the calculation in other cases.

Finally, I'm concerned about the use of the proportional error model:

Y=F*(1 + ERR(1))

This error model is inappropriate when the prediction F may be zero, as will happen for very large values of time when both EINA and EINB become zero. It is necessary to use the additive model Y=F+ERR(1), or a combined additive and proportional model.

-- Alison Boeckmann

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