# Re:Re: simluation Tmax Cmax AUC

From: yhb5442387
Date: Thu, 22 Apr 2010 23:49:14 +0800 (CST)

Dear Sebastien:I have to say that my admiration to you is beyond the words,because you have directly pointed out the radical essence.Could the ke or the V/F,any one , be derived from the known Cmax,AUC,?As far as I have leaned ,the equations are as follows: AUC=Dose*F/Ke*V,Cmax=(F*Dose/V)*(exp(Ka*Tmax)-epx(Ke*Tmax)).After the computation and transformation ,I was obstructed in the front of the following: Ke-2.1*Ln(Ke)=3.18.something like that, well I could not remember the constant number (2.1 e.g.) clearly. I am not able to solve that equation.So,I decide to make some equations in the \$PK,in which the CL V and Ka could all be expressed as the component of Tmax ,Cmax,and AUC at the same time.The suggestion you have made is usefull,but still a little problem :the Ke.Thank you for your generous help.22 03:24:55，"Sebastien Bihorel" <Sebastien.Bihorel
>
>It looks like you are trying to derive the unknown values of KA, CL/F
>and V/F from the known values of Tmax, Cmax and AUC (probably obtained
>from a non compartmental analysis).
>
>If that is really the case, you will need to know the values of the
>half-life of elimination or the slope of elimination Ke to solve for Ka
>and V/F (the following solutions assume a one-compartment model with
>linear absorption and elimination):
>
>CL/F = Dose/AUC
>
>V/F = (CL/F)/Ke
>
>KA = - (1/tmax)*ln[ exp(-ke*tmax) - (Cmax*V)/(F*dose) ]
>
>If you don't know Ke, you will have to solve for V/F and Ka numerically,
>
>Sebastien
>
>yhb5442387 wrote:
>>
>> Dear Nmuser:
>>
>> I would like to do a data simulation .The distributions of the drug
>> population pharmaockinetics were assumed to be normal.But the profile
>> that I have got is Tmax, Cmax,and AUC(the means and variances are
>> should I have to do in the model where the advan2 has been taken in
>> the \$subroutine.
>>
>> F=1.
>>
>> Any suggestion would be appreciated.
>>
>> Yours,ye hong bo
>>
>>
>>
>>
>> --
>> 工作和生活,都要开心的过.
>> 你好,叶红波在此送上真挚的祝福.祝你开心,
>> 叶红波
>>
>>

```--

叶红波```
Received on Thu Apr 22 2010 - 11:49:14 EDT

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