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Dosing Regimen Design. Multiple Dosing: Intermittent or multiple dose regimen. 100 mg q. t 1/2 via i.v. bolus. 200. 150. Amount in Body [mg]. 100. 50. Time [t 1/2 ]: 1 2 3 4 5 6. Principle: 1 dose lost per .

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Dosing regimen design

Dosing Regimen Design

Multiple Dosing:

Intermittent or multiple dose regimen


100 mg q t 1 2 via i v bolus
100 mg q. t1/2 via i.v. bolus

200

150

Amount in Body [mg]

100

50

Time [t1/2]: 1 2 3 4 5 6


Principle 1 dose lost per
Principle: 1 dose lost per

At steady state, Rate In = Rate Out

F•Dose = Ass,max - Ass,min

Ass,min = Ass,maxe-KE

F•Dose = Ass,max (1 - e-KE)

Ass,max = F•Dose /(1 - e-KE)

Ass,min = Ass,max - F•Dose


A n max a n min
AN,max & AN,min

N = 1 2 3 4 5 6 7

Not AN,min = AN,max - F•Dose Why?


C ss max c ss min
Css,max & Css,min


Average amount of drug in the body at steady state
Average amount of drug in the body at steady state

At steady state, Rate In = Rate Out

FDose/ = KE Ass,av

1/KE = t1/2/ln 2 = 1.44 t1/2



AUC

Equal Areas


C ss av
Css,av

Concept applies to all routes of administration and it is independent of absorption rate.


Dosing rate from auc 0
Dosing rate from AUC0-

Given AUC after a single dose, D, the maintenance dose, DM , is:

Where D produced the AUC0-, Css,av is the desired steady-state plasma concentration, and  is the desired dosing interval.


Example 50 mg p o dose
Example: 50 mg p.o. dose

2 3 2.4 1.8

Cp

mg/L

1 2 4 8

Time [h]

AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L


Example continued
Example, continued

Calculate a dosing regimen for this drug that would provide an average steady-state plasma concentration of 15 mg/L.

DM = 193 mg

Dosing regimen: 200 mg q. 6 h.


Example continued 2
Example, continued - 2

There is a problem with this approach ??

Peak and trough concentrations are unknown.

Css,max

Css,min


Another example digitoxin
Another example - digitoxin

t1/2 = 6 days; usual DR is 0.1 mg/day

Assuming rapid and complete absorption of digitoxin,

Ass,av = 1.44 F Dose t1/2/

= (1.44)(1)(0.1mg/d)(6d)/(1d)

= 0.864 mg

What would be the average steady-state body level?

Maximum and minimum plateau values?

Ass,max = 0.1/(1-e-(0.116)(1))

= 0.909 mg

Is there accumulation of digitoxin?

Ass,min = 0.909 – 0.1

= 0.809 mg

How long to reach steady state?


Rate of accumulation ai fi

KE

AI

KE

FI

Rate of Accumulation, AI, FI

The rate of accumulation depends on the half-life of the drug: 3.3 x t1/2 gives 90% of the steady-state level.

Accumulation Index (AI): Ass,max/F DM = (1 – e-KE)-1

Fluctuation Index (FI): Ass,max/Ass,min = e+KE

When  = t1/2

AI = FI = 2


Absorption rate influence on rate of accumulation

ka

v

CL

Absorption Rate influence on Rate of Accumulation

KE = 0.1


When ka >> KE, control is by drug t1/2:

When ka << KE, control is by absorption t1/2:


Loading dose ld
Loading Dose (LD)

= Ass,max F

Whether a LD is needed depends upon:

  • Accumulation Index

  • Therapeutic Index

  • Drug t1/2

  • Patient Need


Dosing regimen design1

Cp

Time

Dosing Regimen Design

OBJECTIVE: Maintain Cp within the therapeutic window.


Dosing regimen design2

Cu

Cp

Cl

max

Time

Dosing Regimen Design

APPROACH: Calculate max and DM,max.


D m max and dosing rate

Cp

Time

DM,max and Dosing Rate

From the principle that one dose is lost over a dosing interval at steady state:

DM,max = (V/F)(Cu - Cl)

The Dosing Rate (DR) is DM,max max


KEV = CL

(Cu - Cl)/ln(Cu/Cl) = Css,av = logarithmic average of Cu and Cl.

The log average is the concentration at the midpoint of the dosing interval; it’s less than the arithmetic average.

DR = (CL/F)Css,av


Average concentration approach
Average Concentration Approach

  • Choose the average to maintain:

    Css,av = (Cu - Cl)/ln (Cu/Cl)

  • Choose :

    max ;usually 4, 6, 8, 12, 24 h

  • Calculate DR:

    DR = (CL/F)Css,av

  • Calculate DM:

    DM = DR•


Example
Example

Css,av = (10 – 3)/ln (10/3) = 5.8 mg/L

max = (1.44)(4.85)[ln (10/3)] = 8.41 h

Choose  < max: 8 h

DR = (5 L/h)(5.8 mg/L)/(0.8) = 36.25 mg/h

DM = (36.25 mg/h)(8 h) = 290 mg  300 mg

Dosing Regimen: 300 mg q 8 h


Peak concentration approach
Peak Concentration Approach

  • Choose the peak concentration to maintain.

  • Choose :

    max ;usually 4, 6, 8, 12, 24 h

  • Calculate DM:

    DM = (V•Cpeak/F)(1 - e-KE)

    from:


Example1
Example

Cpeak = 8 mg/L

max = (1.44)(4.85)[ln (10/3)] = 8.41 h

Choose  < max: 6 h

 set to 6 h so that Css,min > Cl

DM = [(35 L)(8 mg/L)/0.8](1 – e-(0.143)(6)) = 202 mg

Dosing Regimen: 200 mg q 6 h


Check
Check

Css,max = [(0.8)(200)/35]/(1 – e-(0.143)(6)) = 7.93 mg/l

Css,min = 7.93 - (0.8)(200)/35 = 3.4 mg/L


Rationale for controlled release dosage forms
Rationale for controlled release dosage forms

  • Compliance vs. fluctuation: when the dosing interval is less than 8 h, compliance drops.

    • For short half-life drugs, either  must be small (2, 3, 4, 6 h), or the fluctuation must be quite large, when conventional dosage forms are used.

    • Use of controlled release permits long  while maintaining low fluctuation.

    • Not generally of value for drugs with long half lives (> 12 h). Due to extra expense, they should not be recommended.


Assessment of pk parameters
Assessment of PK parameters

CL:

CL/F = (DM/)/Css,av and Css,av = AUCss,/

Relative F:

CLR:

CLR = (Ae,ss/ x Css,av) where Ae,ss is the amount of drug excreted in the urine over one .


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