NONMEM Users Network Archive

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BSV and BOV interaction

From: Jia Ji <jackie.j.ji>
Date: Fri, 18 Dec 2009 18:32:39 -0500

Dear All,

I am trying to model our data with a two-compartment model now. In our
trial, some patients received escalated dose at the second cycle so they
have one more set of kinetics data. So there were BSV and BOV on PK
parameters in the model. Objective function value is significantly improved
(compared with the model not having BOV) and SE of ETAs are around 40% or
less. The code is as below:

$PK
  DESC=1
  IF (TIME.GE.100) DESC=2
  IOV1=0
  IF (DESC.EQ.1) IOV1=ETA(2)
  IF (DESC.EQ.2) IOV1=ETA(3)

  IOV2=0
  IF (DESC.EQ.1) IOV2=ETA(5)
  IF (DESC.EQ.2) IOV2=ETA(6)

  ETCL = ETA(1)+IOV1
  ETQ = ETA(4)+IOV2
  ETV2 = ETA(7)

  CL=THETA(1)*EXP(ETCL)
  V1=THETA(2)
  Q=THETA(3)*EXP(ETQ)
  V2=THETA(4)*EXP(ETV2)

;OMEGA initial estimates
  $OMEGA 0.0529
  $OMEGA BLOCK(1) 0.05
  $OMEGA BLOCK(1) SAME
  $OMEGA 0.318
  $OMEGA BLOCK(1) 0.05
  $OMEGA BLOCK(1) SAME
  $OMEGA 0.711

When I looked at scatterplot of ETA, I found that there is strong
correlation between ETA(1) and ETA(2), which is BSV and BOV of CL. And the
same thing happened to BSV and BOV of Q. Worrying about
over-parameterization (I am not NONMEM 7 user), I tried to define a THETA
for this correlation as the code below (just test on CL only first):

$PK
  DESC=1
  IF (TIME.GE.100) DESC=2
  IOV1=0
  IF (DESC.EQ.1) IOV1=THETA(1)*ETA(1)
  IF (DESC.EQ.2) IOV1=THETA(1)*ETA(1)

  ETCL = ETA(1)+IOV1
  ETQ = ETA(2)
  ETV2 = ETA(3)

  CL=THETA(2)*EXP(ETCL)
  V1=THETA(3)
  Q=THETA(4)*EXP(ETQ)
  V2=THETA(5)*EXP(ETV2)

The objective function value is exactly the same as the model not having
IOV. BSV of CL is decreased and SE of THETAs are also improved, though. The
same thing happend to Q when tested individually. Then I tried another way
to account for this correlation:

$PK
  DESC=1
  IF (TIME.GE.100) DESC=2
  IOV1=0
  IF (DESC.EQ.1) IOV1=ETA(2)
  IF (DESC.EQ.2) IOV1=ETA(3)

  ETCL = ETA(1)+IOV1
  ETQ = ETA(4)
  ETV2 = ETA(5)

  CL=THETA(1)*EXP(ETCL)
  V1=THETA(2)
  Q=THETA(3)*EXP(ETQ)
  V2=THETA(4)*EXP(ETV2)

;OMEGA initial estimates
  $OMEGA BLOCK(2) 0.0529 0.01 0.05
  $OMEGA BLOCK(1) 0.05 ;BTW, I don't know how to do SAME here, it's
not working when putting SAME here
  $OMEGA 0.318
  $OMEGA 0.711

This time I got significantly decreased objective function value, compared
with the model not having IOV. But, SE of ETA(1), ETA(2) and ETA(3) are
huge!

All together, does it mean that there is no need to have BOV on CL and Q? Or
I don't get the right solution to solve correlation problem? Any suggestion
is highly appreciated! Thank you so much!

Happy Holidays!

Jia

Received on Fri Dec 18 2009 - 18:32:39 EST

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