Dosing Regimen Design

Dosing Regimen DesignMultiple Dosing:

Intermittent or multiple dose regimen

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

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

Amou

nt in

Bod

y [m

g]

50

100

150

200

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

AN,max & AN,min

E

E

ENK

ssK

NK

N eADoseFeeA

1

11

max,max,

N = 1 2 3 4 5 6 7

EKNN eAA max,min,

Not AN,min = AN,max - F•Dose Why?

Css,max & Css,min

EKss eVDoseFC

1max,

VDoseFCeCC ss

Kssss

E

max,max,min,

Average amount of drug in the body at steady state

At steady state, Rate In = Rate Out

FDose/ = KE

Ass,av

2/1

,44.1 tDoseF

KDoseFAE

avss

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

Average steady-state plasma concentration

avsCCLDoseF,

CLDoseFC avss,

AUC

Equal Areas

AUCAUC 0

0

,AUCC avss

Css,av

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

Dosing rate from AUC0-

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

DAUCC

D avssM

0

,

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. doseCp

mg/L

1 2 4 8Time [h]

2 3 2.4 1.8 1012

201

0

AUC

5.2122

322

1

AUC

4.5242

4.234

2

AUC

4.8482

8.14.28

4

AUC

63.08.18

Elast KCAUC

AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L

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

mgLhmghLmgD

AUCC

D avssM 50

/3.236/15

0

,

DM = 193 mg

Dosing regimen: 200 mg q. 6 h.

Example, continued - 2There is a problem with this approach ??Peak and trough concentrations are unknown. Css,max

Css,min

Another example - digitoxin

t1/2 = 6 days; usual DR is 0.1 mg/dayAssuming rapid and complete absorption of digitoxin,What would be the average steady-state body level?

Maximum and minimum plateau values?

Is there accumulation of digitoxin?

Ass,av = 1.44 F Dose t1/2/= (1.44)(1)(0.1mg/d)(6d)/(1d)= 0.864 mgAss,max = 0.1/(1-e-(0.116)(1)) = 0.909 mgAss,min = 0.909 – 0.1 = 0.809 mgHow long to reach steady

state?

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

AI

FI

KE

KE

When = t1/2

AI = FI = 2

Absorption Rate influence on Rate of Accumulation

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

0 20 40 60 80 100 120

Time

ka=0.5.pwo

ka=0.01.pwo

v

CL

ka

KE = 0.1

Ea

NKa

Ea

NkE

ssss

N

Kkek

KkeKf

AA Ea

1max,

max,

Ea

NKa

Ea

NkE

ssss

N

Kkek

KkeKf

AA Ea

1max,

max,

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

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

ENKss ef 1

aNkss ef 1

Loading Dose (LD)Whether a LD is needed depends upon:

•Accumulation Index

•Therapeutic Index

•Drug t1/2

•Patient Need

= Ass,max F

Dosing Regimen DesignOBJECTIVE: Maintain Cp within the therapeutic window.

Cp

Time

Dosing Regimen DesignAPPROACH: Calculate max and DM,max.

Cp

Time

Cu

Clmax

l

u

E

l

u

Kul

CCt

KC

CeCC E

ln44.1ln

2/1max

max

DM,max and Dosing RateFrom 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

lu

luE

Elu

luM

CCCCFVK

KCCCCFVD

DRln/lnmax

max,

Cp

Time

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

lu

luE

Elu

luM

CCCCFVK

KCCCCFVD

DRln/lnmax

max,

Average Concentration Approach1. Choose the average to maintain:

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

2. Choose : max ;usually 4, 6, 8, 12, 24 h

3. Calculate DR:DR = (CL/F)Css,av

4. Calculate DM:DM = DR•

ExampleV = 35L KE = 0.143 h-1 t1/2 = 4.85 h Cu = 10

mg/LCL = 5L/h F = 0.80 Cl = 3 mg/L

Css,av = (10 – 3)/ln (10/3) = 5.8 mg/Lmax = (1.44)(4.85)[ln (10/3)] = 8.41 h Choose < max: 8

hDR = (5 L/h)(5.8 mg/L)/(0.8) = 36.25 mg/hDM = (36.25 mg/h)(8 h) = 290 mg 300 mgDosing Regimen: 300 mg q 8 h

Peak Concentration Approach

1. Choose the peak concentration to maintain.2. Choose :

max ;usually 4, 6, 8, 12, 24 h3. Calculate DM:

DM = (V•Cpeak/F)(1 - e-KE)

from: EKss eVDoseFC

1max,

ExampleV = 35L KE = 0.143 h-1 t1/2 = 4.85 h Cu = 10

mg/LCL = 5L/h F = 0.80 Cl = 3 mg/L

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 EKss eVDoseFC

1max,

VDoseFCC ssss

max,min,

Css,max = [(0.8)(200)/35]/(1 – e-(0.143)(6)) = 7.93 mg/lCss,min = 7.93 - (0.8)(200)/35 = 3.4 mg/L

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 parametersCL:CL/F = (DM/)/Css,av and Css,av = AUCss,/Relative F:

B

M

A

M

Aavss

Bavss

A

B

D

D

CC

FF

,,

,,

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